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阿基米德(Archimedes)是古希臘三大數學家和歷代三大數學家之一,畢生對數學貢獻無數,在二千年前已著手研究曲線的周長、面積和體積問題,其獨特的思考方法開創了許多個數學概念的先河,包括積分、三次方程解和結合力學到數學研究方法上。

今次有幸選擇介紹阿基米德作為功課題目,在參考多本權威史記後,筆者以探討阿基米德的數學成就和研究方法兩部分來寫成本文,而且列出部分命題的證明,希望讀者除了認識這位傳奇人物的偉大之處,亦可嘗試在他的思想世界內穿梭。

他的著作繁多,每本也針對一些數學問題而寫成,在討論那些問題時,往往引伸其他有趣的數學問題,促成他在多個數學分支的成就。因此,在數學成就的部分,本文將對各著作和各成就作詳細敘述。

研究方法的部分是本文的精粹,那裡把數學成就的各章融為一體,析述阿基米德在研究背後的數學哲學,包括利用非邏輯方法探索數學問題。

作為介紹數學歷史的文章,無可避免談及一些數學名詞和概念,有時更可能魯莽地說得含混不清,筆者已盡有限的能力,把不容易明白的地方(縱然可能存在無限個)闡述清楚,務求無論是擅長與不擅長、喜歡與不喜歡數學的人也看得懂這篇文章。

阿基米德(Archimedes)生於西元前 287 – 212 年,出身於西西里島(Sicily)的敘拉古(Syracuse),是天文學家費狄(Pheidias)的兒子,據說與敘拉古的國王希倫二世(Hieron)有親戚關係。

處身古希臘文化最鼎盛的時代──亞歷山大時代,年輕時的阿基米德曾到亞歷山大,跟隨歐幾里德的門徒學習,結交了科農(Conon)、厄拉多塞(Eratosthenes)等好友,回到敘拉古後仍跟阿歷山大的學者保持聯絡。

阿基米德最為人所熟悉的是在力學的貢獻,這可能與他的傳奇事蹟有關 [3]。他曾說過,只要給他一個點,便能舉起地球,說的正是槓杆原理,然而他的確曾利用槓杆原理,憑個人之力移走笨重的大船。又有一次,他在浸浴時發現了物件在液體浮動的原理,高興得裸著身體走到街上,大喊「Eureka!」,即是「想出來了!」的意思。

最受人欽佩的是,他在國家的危難之秋,放棄本來的研究工作,利用力學製成許多威力強大的武器,力抗羅馬人入侵,歷史上記載了敵軍一句說話:「他(阿基米德)從容地坐在海邊,把我們的船像擲錢遊戲似的拋來拋去……還射出那麼多發彈,比神話裡的百手妖怪還厲害。」([6],頁 68)後來羅馬軍偷偷潛入軍營殺死阿德米德,一代宗師就此為國捐軀。

事實上,阿基米德在數學上的成就不下於力學方面。他既富創意又能兼顧嚴謹性,對於許多具有相當難度的核心問題,毫不畏懼地提出新穎的觀點,而且利用精密的邏輯演譯,把它們具體地呈現於人前,這些都是受後世推崇備至,與歐幾里德和阿波羅尼(Apollonius of Perga)被譽為古希臘三大數學家、與牛頓和高斯合稱史上三大數學家的原因。

阿基米德撰寫過多部著作,根據《阿基米德全集 [6] 所收錄,這包括《論球和圓柱》、《圓的度量》、《論劈錐曲面體與旋轉橢圓體》、《論螺線》、《論平面圖形的平衡》、《沙粒的計算》、《拋物線求積法》、《論浮體》、《引理集》和《方法》,每本都針對某種數學或物理問題而寫成,而且在研究過程中往往引伸其他有趣的數學問題,因此他的成就遠超這些書名指明的範圍。以下只討論他在數學的成就,而且為了突顯在各數學分支的貢獻,成就的分類不一定以著作為單位。

2.1 球和圓柱

對球體和圓柱的研究是阿基米德最傑出的數學成就之一,雖然對圓周率所知不多,而且礙於數字觀念還未完整,不能在計算比例時使用無理數,他仍然指出許多體積之間和面積之間的比例關係,令球體體積和球體表面面積公式呼之欲出。有關的資料主要記載在《論球和圓柱》第 I 和第 II 卷中。

2.1.1 《論球和圓柱》

他以六個定義和五個假設開始寫作,其中第五個假設後來成為著名的阿基米德公設(Axiom of Archimedes):「任何兩線段 ab ,如果 a < b ,則必存在正整數 n,使得 na > b 成立。」全卷共有 44 個命題,如他給多西修斯(Dositheus)的信上所說,主要結果有以下三個。

2.1.1a 命題 33:「球面積等於最大圓面積的 4 倍。」

命題 34 說明球體體積的情況:「球體積是以它的大圓為底、半徑為高的圓錐體的 4 倍。」這看來是命題33在體積方面的推廣,但根據他的另一著作《方法》([4],頁 339 – 340),命題 33 是由命題 34 聯想出來的,顯示書中命題的次序不代表它們被發現的先後。

2.1.1b 命題 42 和 43:「球缺的曲面面積等於一個圓的面積,該圓的半徑等由球缺曲面的頂點到達球缺底部圓周的長度。」

命題 42 和 43,分別討論球缺小於和大於半圓的情況。

2.1.1c 命題 34 的推論:「以球的大圓為底、球的直徑為高的圓柱體,體積是該球體的 。」

這可能是阿基米德最感到自豪的發現,他在生時已指定把這個定理刻在墓碑上,後來雖然被外族羅馬人入侵國家時殺掉,但羅馬人仍依照這個意願把他埋葬。

2.1.1d 命題的演譯

正如以上所說,卷I的命題都是由五個假設出發,承先啟後地逐一導引。使用第二假設,即「同一平面上,任何兩條有公共端點的線(曲線或折線)中,如果不相等和凹向同一方向,並且其中一條是要麼整個包含在另一條內,要麼一部分包含於其中,一部分重合,那麼被包含的線便是兩線中較短的。」,馬上得到命題1:「外切於圓的多邊形的周長大於圓的周長。」即 PA + AQ 長於弧 PQ

命題 2 是「給定兩個不等的量,則可求得兩不等的線段,使得大、小兩線段之比小於大、小兩量之比。」篇幅所限,命題 2 的證明恕未能提供,但最重要的是,結合命題 1,很容易便得到命題 3:「給定兩不等量和一圓,則可作出圓的外切和內接多邊形,使得外切多邊形的邊長與內接多邊形邊長之比小於大、小兩量之比。」由於兩個不等量是任意的,所以亦可任意逼近圓的邊長,從而證明有關圓面積的定理。這種方法叫「窮竭法」,我們將在介紹阿基米德的研究方法時詳加解釋。

其他的命題也是循著差不多的方向,有的是以描述球體體積的版本出現,證明有關球體的定理。後來加入討論扇形和球缺。

2.1.2 《論球和圓柱》II

共九個命題,其中有六個是問題,以第一章為依據而給出,主要的研究對象是球缺和作圖的方法。以下列出那六個問題:

  1. 命題 1:「給定一個圓錐或圓柱,求一球等於該圓錐或該圓柱。」

  2. 命題 3:「用一平面截給定的球,使得兩球缺面之比為已知比。」

  3. 命題 4:「用一平面截給定的球,使得兩球冠體積之比為已知比。」

  4. 命題 5:「作一球冠與一個球冠相似,而與另一個球冠體積相等。」

  5. 命題 6:「已知兩個球冠,求第三個球冠,使其與一個球冠相似,而與另一個球冠有相等的面。」

  6. 命題 7:「用一平面從已知球截出一個球冠,使得該球冠與一圓錐有已知比,此圓錐與球冠同底等高。」

其中命題 4 具有相當的歷史價值,因為它首次引起人類對三次方程的關注,我們將留待在三次方程的部分深入探討。

幾乎任何的數學文明也離不開對圓的研究,阿基米德不遑多讓,而且跟以上一章的討論一樣,再次向神秘的圓面積和圓周率挑戰。關於圓的度量,他寫了一本名為《圓的度量》的著作,但不少地方已經失傳([2],頁 50),現只遺下 3 個命題,但已經顯出了這位數學大師的非凡功架。

2.2a 圓面積計算

命題 1 開宗明義說明「圓的面積等於一個以其周界及半徑作兩個直角邊的直角三角形的面積」。《九章算述》曾說過:「半周半徑相乘得積步」([6],頁 80)。兩者也指出圓面積的計算方法,但阿基米德的發現可能比較中國方面早。

阿基米德的證明如下。設 A 為圓面積、C 為圓周、T 為命題所述的三角形的面積,假若 A > T,我們可作邊數足夠多的內接正多邊形 P 使

而得出 P > T

但這是不可能的,因為把多邊形分割成大小一樣的三角形,h 比半徑 r 短,而 P 的周界亦比 C 短,所按照計算面積的方法,P < T,與以上所說矛盾。同理,我們知道 A < T 也不成立,所以 A = T。這種證明方法在今天也十分常見,叫做「歸謬法」,我們將在研究方法的部分繼續討論。

2.2b 圓周率

歐幾里德在《幾何原本》討論了許多圓的等性,但沒有提到圓周率的值和圓面積、圓周的計算方法([6],頁 81)。阿基米德卻在科學史上,首創使用上下界來解定一定量的近似值,而且提供了誤差的計算([6],頁 81)。這些都記錄在命題 3 ──「任何一個圓周與它的直徑的比小於 而大於 」。

如圖,

(),

(沒有人知道他何以寫出 的近似值,我們將在討論算成就時詳加分析)

兩式相加,得

我們可容易地證明 ,因而 DOBC 的面積可分別等於 和 ,於是有

左右同時加 1,

左端分母與右端分子交換,再由前面的不等式,有

或 。

於是我們證明了正六邊形與直徑比例的上界,以相同方法,阿基米德計算正十二邊形至正九十六形的比例,亦以差不多的方法計算應對的下界,最後得出命題中的上下界。

阿波羅尼(Apollonius of Perga)是與阿基米德同期的數學家,他以研究圓錐曲線名垂千古,研究時所用的座標方法,更令後世的費馬(Fermat)得到啟發而創立解析幾何([8])。相比之下,阿基米德對拋物線、拋物線體、雙曲線體和橢圓球體的研究不如前者聞名,但阿基米德運用力學和比例來探討問題的獨特風格([4],頁27 – 34、332 – 338),仍贏得後世不少注視。這裡他大量採用了窮竭法來證明命題,但我們將在稍後才討論研究方法,這裡只略述他的研究成果。

他寫過一本名為《拋物線求積法》的著作,有 24 個命題,最重要的是命題 24──「由任一拋物線弓形的面積是同底等高的三角形面積的 。」

他又寫過另一本著作名為《論劈錐曲面體與旋轉橢圓體》,共有命題 32 個。劈錐曲面體即是由拋物線或雙曲線兩種圓錐曲線,迴轉所形成的立體。旋轉橢圓體也是一樣,只是圓錐曲線是橢圓形。阿基米德的研究集中在體積問題。以下是較重要的兩個命題:

  1. 命題 21 和 22:「旋轉拋物面任一截段的體積是與其同底同軸的圓錐或圓錐截段的體積的 。」
  2. 命題 24:「在迴轉拋物面體上,以兩平面切割(任意方式),則分成的兩截段(拋物面體)為它們軸長平方之比。」

    截段 APp : 截段 APp

    = AN2 : AN2

古典希臘時期對幾何的研究多限制於能繪製的圖形內,外形奇特的曲線往往被忽視,直至阿歷山大時代才有人打破這種思想,阿基米德便是其中一人([5],頁 125)。他在《論螺線》中討論一種新奇的曲線──一直線以一端點(稱為原點)為中心,作均速旋轉,直線的另一端點以均速向外延伸。這種曲線稱為「螺線」(spiral),後來被稱為「阿基米德螺線」,用極方程表示,即 r = aq

沒有人知道他何以繪畫這種曲線,最多相信的說法是他利用了運動學的原理,把兩種速度以向量的形式結合起來,從而畫出螺線,這是史上首個微分概念([1],頁 150),第二次出現要直到 1629 年的費馬(Fermat)。然而,阿基米德竟於《論螺線》中提出了在螺線上作切線(tangent)的技巧,他的思維就是這麼匪夷所思。

《論螺線》有 28 個命題,前 9 個是關於圓和切線的比例關係,命題 10 至 12 討論算術級數,其中包括

為後來討論螺線的面積問題鋪路。命題 13 至 20 研究螺線的切線,命題 20 帶出了切線的作圖方法:如果 P 是第一圈上的任意一點,作 OP 的垂線 OT,那麼過 P 點的切線將與 OT 交於 T;如果以 O 為圓心,OP 為半徑的圓交起始線於 K,那麼 OT 等於該圓上 KP 間在前段方向與弧長。接著還提出在任意第 n 圈的處理方法。

由命題 21 開始講述面積,證明方法亦是窮竭法,其中命題 21 指出:

若螺線是 r = aq,則由螺線第一圈和始線所圍圖形的面積 = 。

古希臘的一些人存有一個誤解,就是世上的沙粒無窮多,即使不是無窮多,也沒有一個可以寫出來的數可以形容。阿基米德希望糾正這種觀念,提出一種新的記數方法來表達數量非常大的數字,而且計算出宇宙中的沙粒總數,並把這項工作收錄在《沙粒的計算》,那是他唯一的算術著作,縱使他在算術還有其他的卓越成就。當然希臘的數字符號中,以代表「萬」的「M」為數值最大,若要表示 2 萬,則用 ,其中 b 代表 2。於是,希臘符最大只可以表示一萬的一萬倍,即一億(108)。新的記數方法利用指數概念,稱由 1 到 108 為第一級數,由 108 到 1016 為第二級,如是者,第 108 級數是由 到 ,以上都稱為第 1 周期的數字,而且用 P 代表 。在第二周期,第一級是由 P.1 到 P.108,如是者,第 108 級即由 P.到 P.。這樣下來,第 108 周期的第一級數是由 .1到 .108,第 108 級數是由 . 到 .。阿基米德假設了地球的大小、與月球的距離、太陽的大小、宇宙的大小和一夥罌粟種子的沙量數目,計算出全宇宙的沙粒數目僅有 1051 ,遠少於 .,從而澄清誤解。阿基米德並未就此滿足,本來可以把記數方法發展至能夠達任意大的數量,但看見目標已達到,而且可能當然臨近羅馬人入侵的時刻,沒有進一步進行改革記數制度的工作([6],頁 90)。

主要成就有兩個,第一是給出 的近似值,第二是對數值大的數字,給出平方根的近似值。

2.6a 的近似值

我們在 2.2b 也說過,阿基米德在尋找圓周率的近似值時提出了 的近似值,實際上是這樣的:

有趣地,這兩個分數分別是分母不大於 153 和分母不大於 780 的分數中,最接近 的([6],頁 81)。究竟他是如何得到這兩個分數呢?從古到今眾說紛紜,較多人相信的是他利用了不等式

原因是他的好友海倫曾利用相似的計算方法給出 的近似值,阿拉伯人阿里卡西(Alkarkhi,公元 11 世紀)吸收了希臘原始資料之後,使用了 ([4],頁 68)。

一般相信,阿基米德以 開始估值,得 ,與 比較,,代入 變成,把它成為 a 的新值,重覆以上步驟,最後有 。將同樣方法應用到 ,於是便成了 。

其實他可以不用 ,改用 作為 a 的初始值,而且以上方法可以重覆更多次數來進一步逼近 ,但他沒有這樣做,相信他認為自己的選擇,不但已足夠接近 ,亦方便在計算中抽取公共因子([4],頁 70),真是照顧周詳。

2.6b 大數平方根的近似值

他主要利用了歐幾里德定理 (a + b)2 = a2 + 2ab + b2 和以 60 為分母的分數。

古希臘數學家偶爾探討三次方程,但只停留在純三次方程(y = ax3)的層面,正如阿基米德在《論球和圓柱》命題 1 和 5 中所處理的 a2 : x = x : b 這類問題。然而,他在命題 4 面對的是這樣一個比例:

當中 a 是球半徑,m : n是給定比(兩直線段之比,m > n),x 是較大一段的高。他在化解過程時把問題變得更一般化,即成為:

當中把線段 a 分成 (ax , a) 兩段,bx 被理解為正方形面積。他藉著求出拋物線和雙曲線的交點來得到答案,還指出在 0 與 a 之間無根、有一根或兩個根的條件([4],頁 116 – 117)。

當時專門研究圓錐曲線阿波羅尼(Apollonius of Perga),在探討「從一定點作某圓錐曲線的法線(normal)的條數問題」時也遇上相似的問題,但他沒有採用三次方程,而是於致力於幾何上的直接解法,阿基米德的研究結果比阿波羅尼更一般和廣泛([4],頁 119)。

阿基米德開闢了一條用圓錐曲線化解三次方程的路,後來傳入阿拉伯國家,引起哈津(al-Khazin,? – 965?)、伊本海塞姆(Ibn al-Haytham,965 – 1040)、艾布里朱德(Abu’l Jud,約 1000)沿著這方向研究,最後奧馬海亞姆(Omar Khayyam,1048 – 1131)總結出所有類型的三次方程的圓錐曲線解法([6],頁 80)。

作為三大數學家之一,阿基米德對數學的研究多不勝數,有一些已經失傳,或者只尋回小部分,亦有一些根本沒有記錄在長篇大論的著作,而是以書信和其他輕便的方式流傳下來,這裡都是比以上所提暗淡一些,但亦歷久不衰的數學成就。

2.8a 三角形面積公式

三角形面積 =,其中 s 是三角形周界之半,abc 是三角形的邊長。

2.8b 半正多面體

2.8c 正七邊形作圖法

2.8d 群牛問題

阿基米德在一次探望敘拉古國王赫尼洛時提出一個問題,意思如下:

Ww 分別是白色公牛和母牛的數量,

Xx 分別是黑色公牛和母牛的數量,

Yy 分別是黃色公牛和母牛的數量,

Zz 分別是花斑公牛和母牛的數量。

然後又對牛隻的數量給出一些條件,用數式表示即:

問各種顏色的公牛和母牛有多少隻。

赫尼洛好不容易找到答案,最小的答案也是七位數。阿基米德在問題加入兩個條件:

赫尼洛再解答不到,數學歷史學家也覺得阿基米德亦無法解答,原因是在 1965 年借助電腦才找到答案,答案的位數多達 206500 個([2],頁 97 – 98)。

無論如何,從解破第一部分的問題和預知第二部分問題的難度,可看到阿基米德的代數造詣之高。

2.8e 《引理集》

有一本叫《引理集》的阿拉伯書流傳下來,記載了 15 個有關圓形的問題,相信是阿基米德的後人替他整理的。因此,《引理集》的命題好像沒有共同目的,也沒有使用窮竭法,只是閒談一些圓的特性。以下是一些例子:

  1. 命題 4:「如果 AB 是某半圓的直徑, NAB 上任意一點,以 ANBN 作直徑得兩個半圓,那麼它們圍成的面積,等於以 PN 為半徑的圓形,這裡 PNAB 的垂線,交半圓於 P。」

  2. 命題 9:「如果 ABCD 不是直徑,且相交成直角,則
    (弧 AD)+(弧 CB)=(弧 AC)+(弧 DB)。」

繼承前人的成果,提出富創意的意念,而且堅持尋找嚴謹的證明都是阿基米德的研究風格([4],頁 26 – 27)。一如廣受認同的幾何原本,他的著作亦是由定義和假設出發,把命理導引出來。以下將介紹他的證明方法。另外,我們從《方法》([4],頁 339 – 340)知道,陳述命題的先後次序,不等於被發現的先後次序,由此肯定阿基米德的探究方法,並不是只有如著作上的邏輯方法,背後還有甚麼秘密呢?以下亦將為你揭開真相。

無論在陳述命題或列寫證明時,阿基米德的著作都充滿著比例,相信在數學成就的部分,大家已看過他把比例操控得如何刁鑽吧。事實上,古希臘的數學主要包含兩種方法,一個是運用面積,另一個是運用比例。歐幾里德兩者也有採用,阿波羅尼(Apollonius of Perga)主要取前者,阿基米德則主要取後者([4],頁 27)。例如圓錐曲線,阿波羅尼以面積來理解 y2,並指出拋物線有 y2 = px 的關係,橢圓和雙曲線有 的關係,當中 p 是變量(parameter),d 是直徑(diameter),而阿基米德則把拋物線視為 ([2],頁 139),因而研究的方向和成果各有不同。

歐幾里得在《幾何原本》第十二卷第二個命題的證明中,使用了一種被人稱為「窮竭法」的比例方法,阿基米德證明自己的命題,亦大量使用了這種方法。我們在 2.1.1d 提過一個窮竭法的例子,先任意指定兩個不同的數值,它們可以形容一個比例 R,並且在一個圓形旁邊,作圓外切多邊形和圓內接多邊形,如果多邊形邊數足夠,它們的圓面積的比例一定比 R 少,阿基米德已證明了這一點。如果 R 是非常接近 1,那麼圓外切多邊形和圓內接多邊形的面積比例亦非常接近 1,這樣便可知道夾在中間的圓形的面積。

這是一種逼近的方法,本來由曲線所圍成的面積是不容易計算的,但利用外切和內接多邊形便可以任意地逼近了。事實上,同樣的方法對圓錐曲線,甚至對由曲線迴轉而成的立體也可以做到逼近的效果,阿基米德利用了這一點,在《論球和圓柱》、《圓的度量》、《拋物線求積法》、《論劈錐曲面體與旋轉橢圓體》和《論螺線》大量應用了窮竭法,衝破了一個又一個難題。

窮竭法不一定需要外切多邊形和內接多邊形同時存在,有時只要有其中一款已經足夠。例如之前提過在《拋物線求積法》的命題 24-「由拋物線與弦 Qq 所圍成的弓形面積等於同底等高三角形面積的 」,證明([5],頁 119 – 120)方法是依靠一系列的三角形來逼近拋物線,先找出頂點 P,作 DPQp,找出 PQPq 之間的頂點 Rr,作DPQR 和 DPqr。事實上,先前的命題證明了

DPQR 的面積 = DPqr 的面積 = DPQp

所以,以上著色部分面積總和 = DPQp + DPQp

阿基米德證明了,在 DPQR 和 DPqr 重覆剛才的步驟,最終可產生一系列的三角形,它們的面積總和與拋物線弓形的面積差可以任意小,因而有

拋物線弓形 PQp = DPQp + DPQp + DPQp + DPQp + … +DPQp

窮竭法告訴我們計算面積的方法或公式,但是這樣還未足以化解問題,因為這些方法和公式需要無窮無盡的運算,當時又沒有無限的概念,要利用間接的方法才可得到答案,這方法通常是「歸謬法」。

「歸謬法」即是反證法-我們先推測答案的數值,然後假設真正的答案大於或小於該數,如果可以證明它們都會產生矛盾,便證明了該數值是答案。回到拋物線面積的例子,阿基米德推測面積是 DPQp,於是假設

由於三角形面積總和與拋物線弓形面積的差可以任意小,所以

拋物線弓形面積 > 三角形面積總和 > DPQp

但根據命題 23〔注〕,若把面積和重新取至 m 項(m > n),

產生矛盾。

同樣地可推出

導致矛盾,於是推測成立。

以上可見,窮竭法和歸謬法是應付曲線問題的黃金拍檔。

看了這麼久,你或許和很多在公元 1906 年前研究阿基米德著作的人有同一個疑問-究竟他如何得到這麼多驚人發現?阿基米德的著作十分簡潔,只記下定義、假設、命題和證明,沒留下絲毫有關如何思考或探索問題的痕跡,讀者只能驚嘆這位天才的功績,卻無法想像這些功績如何達成。作為正文最後一個部分,我們現在就把謎團解開。

1906 年,海格伯(J. L. Geiberg,1854 – 1928)在土耳其發現一本羊皮紙書,裡面記載大量阿基米德的著作,其中包括《方法》,正是阿基米德為了交代自己探索問題的方法而寫成的一本書,全書有 15 個命題,大部分其他著作已出現過,但這裡說明的,不再是嚴謹的證明,而是比證明更有價值的東西。

眾所周知,阿基米德在力學方面亦享負盛名,但原來他會運用力學來研究數學。我們在窮竭法說過的拋物線面積,正是《方法》的命題 1,交代了他寫出證明前的發現:

首先講講上圖各線段的關係。設 BD 為拋物線截段的高(diameter),CF 為在 C 點的切線, P 是在拋物線截線段上的任意點,AKFOPNM 平行於 BDCBNKH 共線,KHKC 長度相同。B 作為頂點,透過之前的定理,EB = BDMN = NPFK = KA。命題 5 又證明了

MO : OP = CA : AO = CK : KN = HK : KN

現在,設想每條線段均有重量,而且重量與長度成正比。如果把 K 視為 HC 的支點,把重量相等於 OP 的線段 TG 放在 HMO 的重心點(中點) N 正在 KN 上,所以支點 K 兩旁保持平衡。

由於 P 是任意的,OP 可以是拋物線截段 ABCAC 之間任何的平行線,所以 MO 可以是 DAFC 內任何的平行線,而且 OPMO 仍保持上一段所說的關係。

阿基米德認為,拋物線截段 ABC 和 DAFC 內的線段數目(OPMO)相同,因而推斷,如果支點仍然是 K,由所有 OP 組成的拋物線截段 ABC,將與由所有 MO 組成的 DAFC 平衡,而 DAFC 的重心是位於 CK 上的 W,那裡 CW = 2WK

因此,

拋物線截段 ABC: DAFC = WK : KH = 1 : 3(∵CK = KH)。

所以,

拋物線截段 ABC = DAFC = DABC。(之前的命題證明了 4 倍的關係)

《方法》的中心思想,是把要計算的未知量,分成許多微小量,通常利用槓杆,令這些微小量與其他微小量成比例關係,後者的微小量的和是較易計算的(例如剛才的三角形 DAFC),從而求出未知量([6],頁 74 -75),這是積分的雛形思想。阿基米德就是這樣把力學和數學結合起來了。

方法固然精彩,但阿基米德指出這是一種直觀的推測,不能代表證明,以後還需要使用其他方法(窮竭法、歸謬法)嚴格證明結果。事實上,力學方法真的不是萬試萬靈,例如考慮下圖([6],頁 75 – 76):

DC 是三角形的高,EF // ABEG ABFH AB,則 GE = HF。隨著 EFAB 承升到 C,在每個高度總找到一對的 EGFH,很容易以為 DACD 和 DBCD 面積相等,其實兩者未必相等。

阿基米德與歐幾里德和阿波羅尼(Apollonius of Perga)同屬阿歷山大時期,合稱古希臘三大數學家,而且他與牛頓和高斯亦被公認為三大數學家。與牛頓的萬有引力一樣,阿基米德最為人熟悉的是力學方面的成就,但既然能在數學方面獲得貫穿古今的聲譽,他對數學的貢獻不容置疑。

阿基米德寫過多部數學著作,包括《論球和圓柱》、《圓的量度》、《論劈錐曲面體與旋轉橢圓體》、《論螺線》、《沙粒的計算》、《拋物線求積法》、《引理集》和《方法》,討論範圍主要在幾何方面,平面的有圓形、拋物線和螺線的切線、周長和面積問題,立體的有圓柱、圓錐、球體、橢圓球體和劈錐曲面的面積和體積問題,還有史無前例地提出圓周率的上下接近值。其他出色的幾何發現包括海倫公式、正七邊形作圖法和半正多面體等,多不勝數。

阿基米德較少討論算術的問題,只有《沙粒的計算》唯一一本算術著作,發明新的記數系統來表達數值非常大的數字。但在研究幾何時,他偶爾碰上一些十分引人入勝的算術問題,先後對 和一些數值很大的平方根提出近似值、發表啟蒙後世的三次方程化解方法、計算算術級數和等,這些仍為不少人津津樂道。

研究方法方面,他是使用比例的天才,在列寫證明和和陳述命題時,比例是他慣用語言。他又喜歡使用力學觀點來探討數學問題,在《方法》的 14 個命題中示範這種獨一無二的思考方法,中心思想正是積分概念。阿基米德深知這種方法不算嚴格證明,因此在其他的著作中一概不露痕跡,改用其他方法證明命題,這時候窮竭法和歸謬法都是他的得力助手。

上述各點使阿基米德成為科學史上,首位提出以上下限來界定一定量的近似值、首位擁有微分概念、首位結合力學和數學。他的風采歷久不衰,至今是所有數學歷史課的必然課題,是所有古希臘歷史書中的天皇巨星。

[1] John Stillwell (1989), Mathematics and Its History, Springer.

[2] T. L. Heath (1965), A History of Greek Mathematics, Volume II, Oxford.

[3] E. T. Bell 著(1937),井竹君等譯(1998),古代學者的近代思想。《大數學家》(Men of Mathematics),頁17 – 34,臺北:九章出版社。

[4] T. L. Heath 著(1912),朱恩寬、李文鉻等譯(1998),《阿基米德全集》(The Works of Archimedes),中國:陝西科學技術出版發行。

[5] Morris Kline 著(1908),林炎全、洪萬生、楊康景松譯(1983),阿歷山大希臘時期:幾何學和三角學。《數學史:數學思想的發展》,第五章,頁 108-125,臺北:九章出版社。

[6] 解恩澤、徐本順主編(1994)。阿基米德《世界數學家思考方法》,頁 61 – 99,濟南:山東教育出版社。

[7] 蔡志強、孫文先(199?)。《數學立體模型製作》。臺北:九章出版社。

玩法

MasterMind 玩法

目標

電腦會隨機抽五種顏色(白、黑、灰、紅、黃、青、藍、紫、啡)。

你要在 12 次之內猜中是那五種顏色和它們的排序。

操控

開始時,電腦會先抽出五種顏色。

用滑鼠按空白處,直至你想要的顏色出現為止。

選好後按「嘗試」。

電腦會告訴你結果:

代表其中一格顏色及位置皆正確

代表其中一種顏色正確但位置不正確

重覆以上步驟,直至完全猜中或用完全部 12 次機會。

數學資料庫 – 數學趣趣地 – 數學文章

歐幾里得的《幾何原本》是一部劃時代的鉅著。在這本書問世以前,人類所累積的數學知識是零碎的。《幾何原本》的重要性在於它將前人的數學知識以邏輯的手法加以組識及整理,使其成為一個嚴密的系統。當然,從生活在二千年後的我們看來,《幾何原本》無疑有其不足和不嚴謹之處,但它研究數學的方法卻被後世數學家一直源用至今。這個專題研究的目的,就是要詳細分析及討論歐幾里得所提倡的這套方法,對《幾何原本》第一卷的定義和公理作深入的探討。更重要的是,在過程中我們學習去了解及欣賞前人的工作。

幾何學是從製造器皿、建築、土地測量等實際問題中產生和發展起來的。石器時代的原始人,已懂得將石器製造成較有規則的幾何形狀。例如在北京西南周口店的猿人遺址中發現的五十萬年前的石器、在山西省襄汾縣丁村發現的幾萬年前的球形石塊均證明了這一點。隨住人類文明的進步,人們不但關心物體的形狀,他們還對物體的大小有具體的要求,這就需要懂得測量及計算長度、面積和體積。這導致一些幾何計算公式的出現。對巴比倫、古埃及或中國古代數學文化有認識的人都知道,這些幾何公式並不像現代的教科書般以「定義—定理—證明」的方式寫出來,而往往是隱藏在例題中。那時的人似乎不重視證明,公式很多時是靠經驗或實驗得出的。正因為這個原因,那些公式有時只能給出一個近似的答案。例如巴比倫人以 A = c2/12(其中 c 為圓周長)作為圓面積的公式,從現代人的角度來看他們取 3 為 p 的近似值。

對幾何學進行全面而深刻的研究,使其成為一門獨立的學科是由希臘人完成的。歐幾里得(Euclid,拉丁文拼為 Euclides 或 Eucleides,希臘文 Eύĸλείδŋs,公元前 300 年前後)的《幾何原本》1更是幾何史上第一本系統地討論幾何的著作。這本書問世以來,在各國廣泛流傳,其影響之大僅次於基督教的《聖經》。這二千年間,千千萬萬的人通過《幾何原本》的邏輯訓練,從而邁進科學的殿堂。牛頓在他的著作《自然哲學之數學原理》的序中寫到:「從那麼少的幾條外來的原理,就能夠取得那麼多的成果,這是幾何學的光榮」(It is the glory of geometry that from so few principles, fetched from without, it is able to accomplish so much)。

直至今天,《幾何原本》無論對數學史或數學教育工作者來說,都有永遠的參考價值。

在《幾何原本》出現之前,許多希臘數學家已做了大量的前驅工作,這些工作大都建立在實驗之上而沒有嚴格的證明。(既使是二千年後的今天,很多重要的數學結果最初出現時都沒有完整的證明。)歐幾里得提倡的公理方法,是一種用來證明命題正確的方法。它不但可以證明命題的對確性,往往可以給出一般的結果。舉個例說,埃及人早就知道以3、4、5為邊長的三角形是直角三角形。希臘人卻可以證明若一個三角形的三條邊長 abc 滿足 a2+b2=c2 的話,那必定是一個直角三角形。若試圖用實驗的方法去驗證這個結論,便需要無限次的實驗,這顯然是不可行的。

到底甚麼是公理方法?試想象你現在要說服其他人命題 P1 是正確的,一個很自然的做法是找出一個你認為大家都會同意的命題 P2,再由 P2 推論出原來的 P1。若大家對 P2 仍有懷疑的話,你便會找出一個比 P2 更簡單的命題 P3,然後由推出 P2……如此類推,直至到達一個人所共知的命題 Pn,你便無需要再解釋了。

倘若你不能達至一個人所共知的命題,你便會陷入一個無限倒退的困境。公理方法的精神就是要避免這種困境,所以我們定下兩條規則:

規則一

接受某些被稱為「公理」或「公設」的命題,對於這些命題我們無需要作任何證明。

規則二

同意何謂「」,即協定一些推論法則。

歐幾里得的成就在於他精心選出 10 條公理,然後從這些公理出發證明了 465 個命題。這些命題當中,有部份(例如畢氏定理)是絕對不明顯的。區區 10 條公理,便反映了歐幾里得對這個世界的理解,便解譯到千變萬化的幾何現象,這確實令人感到驚嘆!

在判斷一個證明的對確性時,除了之前提供的兩條規則之外,其實還有一條:

規則三

在證明之中所用到的字、詞、符號,其內容都應該是大家所清楚知道的。

若果你在證明中用到一個新的名詞,或對舊的名詞有新的用法,你便需要在證明之前先引入一個定義。定義基本上是可以任意給出的,但在大多數的情況下一個定義不應(與自身、其他定義,或公理)產生矛盾。當然,即使一個定義沒有產生矛盾,它是否合理和有用是另一回事。

《幾何原本》的第一卷中有 23 個定義,我們從 [3] 中節錄出較重要的幾條:

15. 是由一條線包圍的平面圖形,其內有一點與這條線上的點連接成的所有線段都相等。

20. 在三邊形中,三條邊相等的,叫做等邊三角形;僅兩條邊相等的,叫做等腰三角形;各邊不相等的,叫做不等邊三角形

23. 平行直線是在同平面內的直線,向兩個方向無限延長3,在不論哪個方向它們都不相交。

我們沒有可能對每一個我們用到的名詞都下定義。要定義一個名詞,必須用另外一些名詞,那些名詞亦需由其他名詞去定義。若果我們不准許出現一些不定義名詞undefined terms),我們將陷入之前遇過的同一個困境:無限倒退。但歐幾里得卻企圖定義所有幾何學中的名詞,「點是沒有部份的」、「線只有長度而沒有寬度」…這些定義是不可能用來做證明的。歐幾里得應該意識到這一點,也許他只是想為討論到的幾何對象給出一個直觀上的描述。

後世對歐氏幾何進行公理化的數學家有很多,其中做得最出色、最直觀與最保留著歐幾里得精神的是19世紀後期20世紀前葉的數學巨人,德國數學家大衛‧希爾伯特 (David Hilbert,1862-1943)。在他的巨著《幾何基礎》4(Grundlagen der Geometrie,1899),規定了幾個不定義名詞:

有了這些不定義名詞,我們便可以定義線段為兩個不同點與其之間的點所組成的集合,可以定義何謂兩點在一線的同側…等等。歐幾里得在寫出第五公設時,便涉及到「線的一側」這個概念,間接假定了線將平面分為兩部份。在《幾何原本》中沒有定義清楚「兩點在線同側」是甚麼意思,亦在沒有解譯的情況下用到「一點在某兩點之間」作論證(歐幾里得覺得那可以從圖中看出),這都是由於歐幾里得未能分清那些概念需要定義、那些應列為不定義名詞所致6。當然,我們不應對《幾何原本》過於吹毛求疵。正如之前所說,歐幾里得推倡的公理方法無可否認是劃時代的突破。當時沒有集合論,歐幾里得沒有理由懂得寫出「線段是兩點與其之間的點所組成的集合」這樣的定義。

在定義 15 裏歐幾里得定義了何謂圓形。歐幾里得所說的「兩線段相等」現時一般有兩種(互相等價的)理解。第一種是「全等」(congruent),有部份作者將「線段的全等」列為不定義名詞(正如「角的全等」一樣),這裏不涉及「長度」(length)的概念;另一種應該是歐幾里得的原意,就是「長度相等」。線段的長度是非常直觀而且自然的概念,亦是歐氏幾何中最重要的一部份(正如引言所說,幾何學源自製造器皿、建築、土地測量等實際問題,均涉及長度和面積)。定義 15 在當時來說可算是非常明確,清楚定義出歐氏幾何中一個重要的成員:圓形(歐氏幾何基本上是研究直線和圓形,或由它們組成的圖形)。可是,後世的數學家不滿意這個定義。甚麼是長度?答案在於希爾伯特的系統中的一個定理:

定理: (線段的長度)

給定線段 OI(稱為單位線段),存在且唯一存在函數 ,它賦予每條線段 AB 一個正實數 ,並滿足

(i) ;

(ii) 當且僅當線段 ABCD 全等;

(iii) 若點 BAC 之間,則 ;

(iv) 當且僅當在 CD 之間存在點 E 使得 ABCE 全等;

(v) 對每個正實數 l 均存在線段 AB 使得 。

這個定理再一次讓我們明白到《幾何原本》中的全部結果均可以建立在一個嚴密的公理系統之上,只是在歐幾里得的年代無法這樣做。縱使《幾何原本》中的證明很多都要依賴於圖,在希爾伯特的公理系統中卻可以將這些定理由公理出發全部證明一次,而不需要依賴於圖,亦不需要另外引入一些貌似明顯的假設。順帶一提,定理中的 (i) 至 (v) 在某程度上告訴我們線段(或直線)應該是連續的。關於連續性的公理亦是《幾何原本》所欠缺的,在下一章我們會討論到這一點。

根據定義 20,等邊三角形不當成是等腰三角形(類似地在《幾何原本》中正方形亦不當成是長方形),現代的教科書一般都不會這樣。把等邊三角形當成等腰三角形在邏輯上是較方便的,因為在證明某個三角形等腰時只需證明其中兩邊相等,而不需理會第三條邊。

定義 23 給出平行線的定義,當中涉及「無限延長」以及「方向」這兩個概念。歐幾里得對這兩個概念到底掌握多少,是值得懷疑的。我們相信,「無限延長」一詞是在翻譯過程中才出現。歐幾里得當時所說的線(直線、曲線)都是有限長,而平行直線的定義原文應該是「無論怎樣延長也不相交」(在下一章我們會討論《幾何原本》的公設,其中一條指出有限直線可以繼續延長)。至於線段可以向兩個方向延長這一說法,歐幾里得對「方向」這個概念的理解依賴於直觀。

歐幾里得在《幾何原本》第一卷的開頭寫下 23 個定義,接下來就是 5 條公設(postulate)與 5 條公理(common notion)。歐幾里得認為公理是適用於一切科學的真理,而公設則只應用於幾何。後人一般都不再這樣區分,將那些不需證明而被接受的命題統稱為公理。在第二章中說過,歐幾里得的成就在於他精心選出的 10 條公理:

公設

  1. 由任意一點到任意一點可作直線。
  2. 一條有限直線可以繼續延長。
  3. 以任意點為心及任意的距離7可以畫圓。
  4. 凡直角都相等。
  5. 同平面內一條直線和另外兩條直線相交,若在某一側的兩個內角的和小於二直角,則這二直線經無限延長後在這一側相交。

公理

  1. 等於同量的量彼此相等。
  2. 等量加等量,其和仍相等。
  3. 等量減等量,其差仍相等。
  4. 彼此能重合的物體是全等的。
  5. 整體大於部份。

這 10 條公理問世以來受到廣泛的討論和批評,特別是對第五公設的研究,使人類明白到歐氏幾何並不是宇宙間唯一的幾何,更導致非歐幾何的出現。在討論這些公理之前,首先指出的是歐幾里得從來沒有天真地假定過這 10 條公理之間沒有矛盾。亞里士多德(Aristotle,公元前 384 至前 322 年)認為我們可定義具有矛盾性質的東西,而公理的真確性應由它們推導出來的結果與現實世界比較以作檢驗。

公設一並沒有指出通過兩點的直線的唯一性,但歐幾里得卻在第一卷的第四個命題中用到。那個命題的內容如下:

第一卷的命題 4:

如果兩個三角形有兩邊分別等於兩邊,而且這些相等的線段所夾的角相等。那麼,它們的底邊等於底邊,三角形全等於三角形,而且其餘的角等於其餘的角,即那些邊所對的角。

《幾何原本》中的證明大致是這樣的:設 ABCDEF 是兩個三角形,兩邊 ABAC 分別等於 DEDF,以及 ÐBAC = ÐDEF。將三角形 ABC「移動」到三角形 DEF 上,使得 ABC 分別與 DEF 重合。到底何謂「移動」姑且不談(歐幾里得當然未能交待清楚!),之後歐幾里得卻說由此可知線段 BCEF 重合。他認為若不重合的話兩線段便會圍出一個區域,這是不可能的。這個證明裏面歐幾里得利用了公設一沒有指出的東西:過兩點的直線(或線段)的唯一性。

公設二犯了和公設一相同的毛病:沒有指出線段的延長線的唯一性,如果可以明確寫出這唯一性會更好。由於當時的希臘人所考慮的線全都是有限長的,有時兩線段必須延長後才會相交,這條公設保證了線段可以延長。

公設三保證了圓形的存在性(其實如同之前兩條公設一樣,歐幾里得強調「構造」多於「存在」),這條公設早在第一卷的命題一便已用到。從現代數學的觀點來看,圓形是與一定點(圓心)有相同距離的點所組成的集合,這樣圓形的存在性由集合論本身的公理所保證,故希爾伯特沒有將這樣一條公設放入他的公理系統裏面。

公設四「凡直角都相等」,初看之下很難捉模歐幾里得的用意。直觀上看,一隻直角可以透過平移和旋轉使其與另一隻直角重合,而公設四保證了在平移和旋轉的過程中「垂直性」不變。更一般地,平移和旋轉並不改變兩直線間的夾角,在之前提及過的《幾何原本》第一卷命題4的證明中亦有間接用到這個事實。歐幾里得倒因果了!事實上應該是命題4(SAS 全等三角形判別法)保證了幾何圖形在平移、旋轉過程中幾何結構不變。希爾伯特看出了這一點,所以將 SAS 判別法8列入公理表當中。

在《幾何原本》的 10 條公理當中,第五公設是最具爭議性的一條。其餘的 9 條公理都非常直觀,而且敘述簡短,相比之下第五公設顯得格格不入。我們有理由相信歐幾里得本人也不太喜歡這條公設,大家請看以下的統計:

從統計表可看出第五公設比其他公設、公理引用得較遲,要到命題 29 才第一次被引用。命題 1 至命題 28 包括一些重要的定理和作圖題,它們的證明有部份比命題 29 更複雜。頭28個命題不需要引用到第五公設的原因不是因為它們簡單,是歐幾里得故意將不需要用到第五公設的命題放到最前,到了命題 29 才迫不得意要用到第五公設!讓我們來看看這個關鍵性的命題 29:

《幾何原本》第一卷命題 29

一條直線與兩條平行直線相交,則所成的內錯角相等,同位角相等,且同旁內角的和等於二直角。

在上圖中,AB // CD,直線 EF 分別交 ABCDGH。《幾何原本》對命題 29 的證明大意是這樣的:若同旁內角不互相,則 EF 其中一側的同旁內角之和少於二直角,於是第五公設指出線段 ABCD 適當地延長後會相交,與 AB // CD 矛盾。

命題 29 是大家熟知的平行線性質,它連同命題 27、28 中的平行線判別定理可以證明很多有用的結果和作圖題,著名的例子是三角形內角和等於二直角。雖然第五公設在《幾何原本》中被引用得很少,但即使單單為了證明命題 29 這個關鍵性的命題,便已值得將它列入公理表中。根據數學史家的考證,《幾何原本》中大部份的結果在歐幾里得之前已經有人知道,但第五公設卻是歐幾里得本人想出來的。歐幾里得看出這個公設的重要性,充分顯示出他的天才!

由於第五公設那樣礙眼,從《幾何原本》問世以來,試圖用其餘 4 條公設(以及 5 條公理)證明第五公設的嘗試就已經開始。很多人確信第五公設可以被證明,可是經過了兩千多年,仍然沒有人能證明出第五公設。正如其他著名的難題一樣,有很多人曾經聲稱自己已經解決了這個問題,但結果無一例外地那些「證明」都暗中引用了不能單靠其餘公理證明的命題。越是對第五公設進行研究,就越令人感到懷疑:到底第五公設能否被證明?這兩千年間數學已發展至煥然一新的樣貌,解析幾何、微積分、微分方程及其他數學分支相繼出現,無數一流數學家在數學界大放異彩,但仍然沒有人能證明到第五公設。法國數學家達朗貝爾(1717 年至 1783 年)在 1759 年說歐幾里得第五公設是「幾何原理中的家醜」!

著名數學家勒人達(Adrien Marie Legendre,1752-1833)是千千萬萬為第五公設著迷的人之一,他花了 29 年的時間研究,發表了一個證明第五公設的嘗試。在討論他的證明之前,首先指出在研究第五公設的過程中,數學家們很早便知道很多與第五公設等價的命題,其中一個是平行公設。

平行公設

直線 L 外有一點 P,則 LP 所在的平面上至多只有一條直線通過 P 而與 L 平行。

平行公設只說通過 P 而平行於 L 的直線最多只有一條,但沒有說過這樣的直線存在,因為存在性可以被證明而無須列為公理。現時有很多教科書都用平行公設以代替歐幾里得的第五公設,其中一個原因是平行公設避開了第五公設中「直線的一側」這個概念。雖然這個概念可以嚴格地定義,但作為一條公理其敘述應越簡潔越好。

勒人達的嘗試(證明平行公設)

P 為直線 L1 外一點,過 PL1 的垂線交 L1Q,過 PPQ 的垂線 L2,於是有 L1 // L2。設 L3 是過 P 且異於 PQL2 的任一直線,我們希望證明 L1L3 相交。

在直線 L3 上取一異於 P 的點 R,使得射線 PR 在射線 PQ 和以 P 為端點並在 L2 上的射線之間。在射線 PQ 的另一側取點 R’ 使得 ÐQPR‘ = ÐQPR。留意到點 Q 在 ÐRPR’ 的內部且 L1Q,故直線 L1 必定與 ÐRPR’ 的其中一條邊相交。若 L1PR 相交,則 L1L3 相交。若 L1PR’ 相交於 A,在射線 PR 上取點 B 使得 PB = PA,於是有(SAS),從而得知 ÐPQB 是直角,亦即 BL1上。這證明了 L1L2 相交。

這個證明是否正確?要回答這個問題便需要細心檢驗證明的每一步。首先,我們應定義清楚證明中每一個用詞,例如「垂直」、「角的內部」、「直線的一側」、「三角形的全等」…等等。然後我們要解釋為甚麼 L1L2 同時垂直於 PQ 就必定互相平行(注意此時不可以用第五公設),要解釋為何點 R’ 存在,要解釋為何 Q 在角的內部,要解釋為何 L1 通過角 RPR’ 的內部的點就必定與 PRPR’ 相交。大家可想而知證明平行公設是多麼困難的一件事!我們不打算在這裏詳細討論勒人達的證明有甚麼漏洞,只是想指出這個證明裏面用到一些不能由第五公設以外其他公理所證明的命題。

經過了漫長的歲月,轉機終於來臨。非歐幾何的發展使研究第五公設的數學家們驚嘆不已。為了證明平行公設,人們先假定與平行公設相反的命題,然後與其他公理一起希望推導出矛盾。這樣的努力無論怎試也不成功,更意外地發展出一些與歐氏幾何截然不同的新幾何學。如果這些幾何不包含矛盾,就表示平行公設(或第五公設)與其他公理獨立,不可能由其他公理得出。人類經過長年累月的經驗,相信歐氏幾何的公理不會推出矛盾的命題。但是,我們又憑甚麼相信這些新幾何亦不包含矛盾呢?1898 年龐加萊(Poincaré)發表了一個見解,認為一個公理地建立起來的結構,如果能給它一個算術解釋,就可以相信它的相容性。因為若果這個結構包含矛盾,那麼算術中亦會出現相應的矛盾。希爾伯特完成了這種「矛盾轉移」,在《幾何基礎》中為非歐幾何給出一個算術解釋,加上人類多年來的經驗使人相信算術是相容的,所以我們應對非歐幾何抱有相同程度的信賴。這個答案已令人相當滿意。

至於《幾何原本》中的 5 條公理,有部份被希爾伯特修改後列入他的公理表中,有些則被取消。有興趣的朋友可參看附錄或有關的書籍,這裏就不再討論了。

最後不可不提的是由於《幾何原本》公理的不足,使歐幾里得無可避免地在證明中不自覺地用到一些貌似明顯的事實。特別是在關於點在直線上的次序、直線和圓的連續性等問題上歐幾里得往往採用直觀的方式作論證,例如在《幾何原本》第一卷命題 1 的證明中,他認為以相異點 AB 為圓心和 AB 為半徑的兩個圓必定相交,這等於暗自假設了圓是連續的。

歐幾里得的《幾何原本》是劃時代的鉅著,雖然它裏面的結果大多數都是前人已知的,但它所採用的公理方法卻被數學家沿用至今。精心選出來的10條公理,充份顯示出歐幾里得的天才及驚人的洞察力。特別是第五公設的引入,吸引了無數一流的數學家嘗試去證明,這導致非歐幾何的出現,令我們對歐氏幾何有更深入的了解。

雖然《幾何原本》的公理系統不全,當中的證明亦有不少缺陷,但它踏出了幾何公理化突破性的一步。從《幾何原本》到《幾何基礎》,單從邏輯上看是一種堵漏補遺,但其實這代表著人類抽象思維的昇華。

[1] Marvin Jay Greenberg, Euclidean and non-Euclidean Geometries: development and history, 3rd Edition, W. H. Freeman and Company. New York, 1994.

[2] Benno Artmann, Euclid—the creation of mathematics, Springer-Verlag New York, Inc., 1999.

[3] 歐幾里德(Euclid)著,藍紀正、朱恩寬譯,歐幾里得‧幾何原本,九章出版社,1996。

[4] 莫里斯‧克萊因(Morris Kline)著,北大數學系數學史翻譯小組譯,古今數學思想(第一冊),上海科學技術出版社,2002 年。

[5] 王懷權,數學的故鄉,王懷權出版,1997 年。

[6] 蔣聲,歐幾里得第五公設,九章出版社,1993 年。

I. 連通公理(Axiom of Connection)

  1. 對於任二點 AB 而言必存在有一直線 a,使 ABa 上。

  2. 任二點 AB 最多只有一直線 a 使得 ABa 上。

  3. 每一直線至少含有二點,至少有三點不在同一直線上。

  4. 不在同一直線上的任三點 ABC 至少有一平面 a,使得 ABC 三點都在 a 上。每一平面至少含有一點。

  5. 不共線的三點最多只有一個平面 a,使得這三個點都在其上。

  6. 若一直線上的二點在一平面 a 上,則該線所有點在 a 上。

  7. 若二平面 ab 有一共同點 A,則必至少還有一個點 Bab 之上。

  8. 至少有四點不在同一平面上。

II. 在其間公理(Axiom of Betweenness)

  1. 若一點 B 在兩點 AC 之間,則 ABC 為不同三點在同一直線上而且 B 也在 CA 之間。

  2. 對於任二點 AB 而言,在直線 AB 上至少有一點 C 使得 BAC 之間。

  3. 一直線上任三點最多其中一點在其餘兩點之間。

    定義

    設一直線 a 上二點 AB。無序對 ABBA 稱為 AB 線段。在 AB 之間的點稱為 AB 的內點,AB 叫端點,其他線上的點(非內點、端點者)稱為外點。

  4. (帕科公理)設 ABC 為平面 a 上不共線之三點,a 上有一直線 a,其中 ABC 都不在 a 上。若 a 過線段 AB 的一點,則 a 必也過線段 AC 或線段 BC 的一點。

III. 全等公理(Axiom of Congruence)

  1. AB 為直線 a 上兩點,若 A’a 上一點(或另一直線 a’ 上一點),則 a(或 a’)上有另外一點 B’,使得此二線段 ABA’B’ 全合,記為 AB = A’B’

  2. AB = A’B’AB = A”B”,則 A’B’ = A”B”
  3. 設直線 aABBC 沒有共同內點,直線 a’A’B’B’C’ 也沒有共同內點且 AB = A’B’BC = B’C’,則 AC = A’C’

  4. 設給定了一平面 a 上的一個角 Ð(h, k),平面上 a’ 的直線 a’,以及指定平面 a’ 上直線 a’ 的一側,且 h’ 是直線 a’ 上一點 O’ 開始的一條射線,則在平面 a’ 上直線 a’ 的指定一側恰有一條射線 k’,使 Ð(h, k) = Ð(h‘, k‘)。每個角與它自己全合。

  5. 兩個三角形 ABCA’B’C’ 若滿足 AB = AB‘、AC = AC‘ 和 ÐCAB = ÐCAB‘,則必有 ÐABC = ÐABC‘。

IV. 平行公理(Axiom of Parallels)

a 為一直線,A 為線外一點,則在 aA 的平面上最多存在一直線過 A 且與 a 平行。

V. 連續公理(Axiom of Continuity)

  1. (阿基米德公理) 若 ABCD 是任意兩線段,則在從 A 開始並通過點 B 的射線上必有這樣的有限個點 A1A2、…、An,使得線段AA1A1A2、…、An-1An都與線段 CD 全合,而且 BAAn 之間。

  2. 設一直線上的點若滿足公理 I(1)、I(2)、II、III(1)、V(1),則不能再擴大成更大的集合使滿足這些公理。

Mathematical Database – Math Funland – Math Articles

The Mathematical Achievements and Methodologies of Archimedes

 

image Volume of the cylinder = ´ Volume of the sphere
Surface area of cylinder (including both ends) = ´ Surface area of sphere

Summary

Archimedes is regarded as one of the three most important mathematicians of ancient Greece. His numerous contributions to the to the study of mathematics also earned him the title of being one of the three most important mathematics ever. It was two thousand years ago that he started his study in problems involving perimeters, areas and volumes. Moreover, his methodologies in thinking ( for example, using concepts of differentials, mechanics and solutions to cubic equations) were unique and pioneering.

The aim of this article is to discuss the mathematical achievements and methodologies of Archimedes, and each of these topics is given a separate section below. The writer hopes that through quoting Archimedes’ proofs to some of his statements, the reader will learn more about the beautiful mind of the mathematician and become better able to understand his thoughts.

In each of his books, Archimedes took a look at some interesting mathematical problems, which in turn led to even more problems. In this way, he made his contribution to may of the branches of mathematics. In the section ‘Mathematical Achievements’, the reader will be given a detailed discussion of Archimedes’ books and achievements. To investigate the philosophy of Archimedes’ studies, the ‘Methodologies’ section will deal with his thinking methodologies. Summarizing the chapters on his mathematical achievements, this section should serve as the center of interest of this article.

As this is an article on mathematical history, there will be some mathematical terms and concepts in the discussion. The writer has tried his best to make the article easy to understand for his readers, whether or not they are keen or good in mathematics.

 

I. The Bibliography of Archimedes

Archimedes (287 – 212 B.C.) was born at Syracuse of Sicily as a son of the astronomer Pheidias. It is said that Archimedes was a relative of Hieron, the king of Syracuse.

Archimedes learnt from the disciples of the mathematician Euclid when he was young. When he was learning at Alexandria, he made friends with Conon, Eratosthenes and many others. He kept in touch with the academics in Alexandria even when he got back to Syracuse.

Archimedes was most famous for his contributions in mechanics. This is probably due to his many amazing stories [3]. He said that he could move the Earth just if he was given a fixed point — and that was his theory of leverage. In fact, he did move a ship all by himself using a lever. Another story is that he discovered buoyancy when he was having a bath. He was so excited about his discovery that he ran naked on to the street, shouting “Eureka!”, which meant “I’ve got it”.

When his country was attacked by Romanians, he put all his studies and researches away and successfully invented a number of powerful weapons. His enemy was quoted to have said, “He (Archimedes) sat at the seashore, and easily threw our boats to and fro as if he was just playing with some coins…..he is even more powerful than the monsters in the legends as he can throw so many bombs at us.” ([6], p.68) in the end, some Romanian soldiers got into Archimedes’ camp secretly and killed him.

However, the achievements of Archimedes in mathematics were no less important. When faced with difficult problems, he gave innovative ideas and logical proofs. That is why he is regarded as one of the three most important ancient Greek mathematicians with Euclid and Apollonius of Perga, as well as one of the three most important mathematicians ever with Newton and Gauss.

 

II. Mathematical Achievements

According to “The Works of Archimedes” [6], Archimedes’ works included “On the Sphere and Cylinder”, “On the Measurement of a Circle”, “On Conoids and Spheroids”, “On Spirals”, “On Plane Equilibriums”, “The Sandreckoner”, “On Quadrature of the parabola”, “On Floating Bodies”, “Book of Lemmas” and “The Method”. Each of his works was focused on some particular problems in mathematics or physics. Very often, he was directed to other interesting problems involved in solving the aforementioned ones. Therefore, his contributions are far greater than the titles of these books might suggest. We will discuss his achievements just in mathematics below. In order to highlight his contributions in the many branches of mathematics, the classification of his achievements may not follow from the works in which the discoveries are noted.

2.1 Sphere and Cylinder

One of the most important of Archimedes’ achievements is his studies on spheres and cylinders. Although there was not much known about the value of the ratio between the circumference and diameter of a circle, and it was impossible to evaluate ratios using irrational numbers with the number system at that time, Archimedes was still able to point out the relationships between the volumes and areas of various geometric shapes. His works nearly gave the exact formulae for volume and surface area for spheres. Related materials can mostly be found in the first two sections of “On the Sphere and Cylinder”.

2.1.1 “On the Sphere and Cylinder”

Archimedes started with six definitions and five hypothesis. His fifth hypothesis later became the famous Axiom of Archimedes, “For any two line segments with length a and b, if a < b , then there exists a natural number n such that na > b .” There were altogether 44 statements in the word. As in his letter to Dositheus, there were three major results, as explained below.

2.1.1a Statement 33: The volume of a sphere is equal to 4 times the area of the largest circle it contains.

image015e

Statement 34 states that, ” the volume of a sphere is 4 times the volume of the cone with height equal to the radius of the sphere and the base identical to the largest circle the sphere contains.”. This seems to imply that statement 34 is an extension of statement 33. Yet, according to “The Method” ([4], page 339 – 340), statement 33 actually came from statement 34. This means that the order in which the statements appeared in the work might not be that of the discoveries.

2.1.1b Statement 42 and 43: The curved surface are of a spherical cap is equal to the area of a circle with radius equal to the distance between the vertex at the curved surface and the base of the spherical cap.image016e

Curved surface area (shaded) = Area of circle of radius r

Statements 42 and 43, discussing the situations where the spherical cap is greater than and less than the semi-circle respectively.

2.1.1c Deduction from statement 34: The cylinder with the large circle of a sphere as base and diameter as height has a volume times of the sphere.

Volume of the cylinder = ´ Volume of the sphere
Surface area of cylinder (including both ends) = ´ Surface area of sphere

This is perhaps the discovery Archimedes felt proudest of: he dictated that this statement to be craved on his gravestone. In fact, the Romanian invaders buried him according to his wish.

2.1.1d Proofs of the Statements

As discussed above, each of the statements in Section I started from five hypotheses. Using the second hypothesis, i.e., “For any two plane curves with common endpoints, if they are not identical and have the same convexity, with one enclosed in the other or partially enclose and partially overlapped, then the enclosed curve is the shorter one”. From this hypothesis we get statement 1, “A polygon inscribing a circle has a larger perimeter than the circle.” This means that PA + AQ is longer than curve PQ.

Statement 2 states that “Given any two unequal numbers, we can find two unequal segments such that the ratio of the lengths of the segments is equal to that of the numbers”. Due to limited space here, the writer will not provide the proof of statement 2. The key point here is that statement 3, which states that “Given two unequal numbers and a circle, we can construct polygons inscribing and inscribed by the circle such that the ratio between the perimeters of the polygons equal to that between the two numbers”, follows easily from statements 1 and 2. As the two numbers are arbitrarily chosen, we can approximate the circumference of the circle to an arbitrarily determined precision, thereby proving statements on the areas of circles. The method of proving statements is called method of exhaustion, and we shall discuss this method in the section “Methods”.

The approaches of other statements are similar. These statements describes the volumes and other properties of spheres, and some properties of sectors and spherical cap.

2.1.2 “On the Sphere and Cylinder” II

There are nine statements, six of which have the form of problems. These problems, based on chapter one, aim to discuss the properties of spherical cap and methods of construction. The problems are listed below:

  1. Statement 1: Find a sphere with an equal volume as a give cone or cylinder.
  2. Statement 3: Construct a plane which divides a given spheres into two spherical cap, such that the ratio of the surface areas is as given.
  3. Statement 4: Construct a plane which divides a given spheres into two curved surfaces of spherical caps, such that the ratio of the volumes is as given.
  4. Statement 5: Construct a spherical cap surface so that it is similar to another spherical cap surface and an equal volume as a third spherical cap surface.
  5. Statement 6: Construct a spherical cap surface so that it is similar to another spherical cap surface and an equal surface area with a third spherical cap surface.
  6. Statement 7: Construct a plane which cuts a given sphere to give a spherical cap surface which has the same base and height with a cone of volume at a given ratio to that of the spherical cap surface.

Statement 4 is of particular historical value as it is the first statement to arouse people’s interest in cubic equations. We will deal with the topic of cubic equations in more depth in a later chapter.

2.2 Measurement of a Circle

Nearly every ancient civilization would came across the problem of circles. On this topic Archimedes wrote a book named “On the Measurement of a Circle”. however, much of the contents of this book was lost ([2], page 50) and only three of the statements remains.

2.2a Evaluation of the Areas of Circles

Statement 1 states that ” The area of a circle is equal to a right-angled triangle with the circumference and the radius as straight edges”. In “The Nine Chapters of Mathematical Art”, there was also a statement meaning ” the product of the semi-circumference and the radius of a circle equals its area” ([6], page 80). Although both statements gave the formula for calculation of a circle’s area, the discovery by Archimedes might be earlier than that in China.

Here is Archimedes’ proof of the statement. Let A be the area of the circle ,C the circumference and T the area of the aforementioned triangle. Assume A > T, we can construct a regular polygon P with a sufficient number of sides such that P is inscribed by the circle and

AP < AT

That is, P > T.

However, this is not possible because we can divide the regular polygon into congruent triangles with h shorter than r, and the perimeter of P is shorter than C implying P < T, contradicting the above. Similarly, we cannot have A < T. And therefore A = T. This method of proving statements, “prove by contradiction’, is now a very common tool. We will discuss more about this in the section “Methodologies”.

2.2b Circumference-Diameter Ratio

In “Elements”, Euclid discussed many properties of circles, but not those involving the evaluation of the ratio between the circumference and diameter, area of circle and length of circumference ([6], page 81). Yet, Archimedes pioneered to use bounding techniques to approximate a quantity and evaluate error terms.([6], page 81). All these works are noted in Statement 3 ── “The ratio of the circumference to the diameter of any circle is between and “.

The proof of Statement 3 is given below ([6], page 82). Archimedes gave an approximation to the value of in his proof.

As in the figure,

(),

.

(Nobody knows how Archimedes get this approximation of , and we will discuss this when we summarize his achievements in arithmetic.)

Adding these two equations, we have

.

It is easy to prove that . Hence, the area of DOBC equals to and respectively. Thus,

.

Adding 1 to both sides,

Rearranging the terms and using the inequality above, we get

or

and this gives the upper bound of the diameter of the regular hexagon. Archimedes also calculated the dimensions of regular 12-gon to regular 96-gon using the same strategy. Similarly, he evaluated the lower bounds and got the bounds in the statement.

 

2.3 Quadrature of the Parabola, Conoids and Spheroids

Apollonius of Perga is a mathematician of the same historical period as Archimedes. He was famous for his studies in conics. His method of coordinates inspired the mathematician Fermat to develop the field of analytical geometry ([8]). Comparing with Archimedes’ studies of parabola, solid bounded by a paraboloid or hyperboloid and a plane, and spheroids, the former is certainly better known. However, the use of proportions and mechanics in Archimedes’ studies was also distinguished ([4], page 27 – 34, 332 – 338). Many of these works were done with the help of the method of exhaustion. We will discuss the results of his studies here and his methodologies later.

There are a total of 24 statements in his work “On Quadrature of the Parabola” and the most influential one is Statement 24── “The area of the parabolic segment of a parabola is equal to times that of the triangle with the same base and height.”

In another book “On Conoids and Spheroids” he gave 32 statements. A conoid is the volume formed by revolving a parabola or a hyperbola. Likewise, a spheroid is formed by revolving an ellipse. Archimedes’ studies focused on the volumes of these forms and here are two of the more important statements in his book:

  1. Statements 21 and 22: The volume of any segment of a paraboloid is times that of a cone (or a segment of a cone) with the same base and axes.
  2. Statement 24: The ratio of the two segments formed by cutting a solid bounded by a paraboloid with two planes in an arbitrary way is equal to that of the squares of the lengths of their axes.
    Segment APp : Segment APp

    = AN2 : AN2

 

2.4 Spirals

A characteristic of ancient Greek geometry studies is the objects were usually limited to those that can be drawn. This implies that some curves with peculiar shapes were often missed. It was not until the Alexandrian period that this ‘rule’ was broken, and Archimedes was one of these rule-breakers ([5], page 125). Hw discussed, in his work “On Spirals”, a special type of curve — the locus of an endpoint of a uniformly-rotating line segment which has the other endpoint (origin) fixed and its length increasing at a uniform rate. This is what we call a spiral or, Archimedes’ spiral. In polar form, the equation of a spiral is r = aq.

It is not known why Archimedes drew this type of curve. A popular suggestion for the reason is that he used the concept of mechanics and combine two velocities vectorically to get the spiral. This is the first concept of differential in history ([1], page 150). The second time this concept appeared was in 1629, introduced by Fermat. Hence, it was indeed amazing that Archimedes could propose in “On Spirals” some techniques of constructing tangents to a spiral.

There are 28 statements in “On Spirals”. The first 9 was about the proportions between circles an their tangents and statements 9 to 12 was on arithmetic progression, including

to facilitate the discussion on area of spirals afterwards. Statements 13 to 20 studies the tangents to spirals. In particular, in Statement 20, a construction method of the tangent was noted: for any point P on the first revolution, construct OT perpendicular to OP. Then the tangent at P would cut OT at T. If we use O as the center and OP as radius, the circle drawn would cut the pole at K, then the length of OT equals to the arc length between K and P. And then he discussed the techniques to use for a point on the nth revolution, for an arbitrary n.

He discussed the matter on area using method of exhaustion from Statement 21 onwards. In Statement 21, he pointed out that:

If a spiral has the form r = aq, then the area bounded by the first revolution of the spiral and the pole is = .

 

2.5 Notations for Numbers

There was a misperception among some ancient Greeks that the number of sand grains in the world was infinite and the number could not be represented by a number. To remove this misperception, in his only work on arithmetic, “Arithmetic on Sand Grains”, Archimedes proposed a new way to denote large numbers and calculated the number of sand grains in the world. At that time, the largest unit in the number system was “M” which meant ten thousands. To represent twenty thousands, the used , where b represented 2. Thus, the largest number that could be represented was ten thousands times ten thousands i.e., 108. The new system used the concept of index: the numbers from 1 to 108 was put in the first class, the numbers from 108 to 1016 the second class, …, and the numbers from to was put in the (108)th class. And all the numbers above were considered to be in the first period. was denoted by P. In the second period, the first class ranged from P�1 to P�108, and the (108 )th class ranged from P� to P�. In this way, the first class in the (108)th period ranged from �1 to �108, and the (108 )th class from � to .. Archimedes hypothesized on the size of the Earth, the distance between the Earth and the Mon, the size of the sum and the Space and the number of sand grains in a seed, and determined that the number of sand grains in the world could just be 1051, which was far less than �. However, the mathematician was not satisfied by these: he could have further developed the system so that it could represent an arbitrarily large number. Yet, having seen that the misperception was gone and probably the Romanians were going to invade the country, Archimedes did not proceed further ([6], page 90).

 

2.6 Approximation of a Square Root

There are two major achievements. One is the approximate value of , and the other was the approximate values of the square roots of some large numbers.

2.6a The Approximate Value of

In 2.2b the writer mentioned that Archimedes found an approximation of when he was studying the circles. Actually, the approximation was:

Interestingly, these two fractions are the closest approximations to with denominators no greater than 153 and 780 respectively ([6], page 81). There were a lot of suggestions to the way by which he obtained these approximations. Many believe that he had used the inequality

.

It is because his friend Heron was known to give an approximation to in this way. Alkarkhi (11th century), an Arabic, used the approximation after consulting some Greek materials ([4], page 68).

Most people believe that Archimedes began the approximation with the value and get . Comparing with , we have . Substituting into we get . Using this as the new value for a and repeating the above process, we get in the end. Using similar techniques, we will get and therefore .

In fact, he could have used instead of as the initial value for a, and he could have done more repetitions to get a better approximation. However, he did not do so. This may be because he thought his approximation was close enough to and it would be easier to extract common factors in further calculations using the above values ([4], page 70).

2.6b Approximate values of some large numbers

In his work “On the Measurement of a Circle”, Archimedes wrote down the following approximations([[4], page 70):

These values are mainly obtained through the Euclid’s theorem (a + b)2 = a2 + 2ab + b2 and fractions with 60 as denominator.

 

2.7 Cubic Equations

Mathematicians in Ancient Greece sometimes discussed cubic equations, but only those in the form y = ax3. This is also reflected by the fact that in Statements 1 and 5 of “Spheres and Cylinders” Archimedes dealt with the equation a2 : x = x : b. However, in Statement 4 he talked about the following ratio:

,

where a is the radius of the sphere, m : n is a given ratio between two line segments with m > n, and x the height of the longer segment. In his solution, he generalized the problem into:

(ax) : c = b2 : x2,

in which the segment a is divided into the two parts (ax , a), and b and x are considered to be the areas of squares. He solved the problem by finding the intersection of the parabola and hyperbola. Furthermore, he pointed out the conditions in which there are zero, one or two roots between 0 and a ([4], page 116 – 117).

Apollonius of Perga also encountered this problem when he was investigating the number of normals that could be drawn through a given point on a conic. Nevertheless, he tried to solve the problem geometrically instead of by dealing with cubic equations. hence, Archimedes’ results offer a more general solution ([4], page 119).

Archimedes’ works in cubic equations inspired al-Khazin (? – 965?), Ibn al-Haytham (965 – 1040) and Abu’l Jud (~1000) to continue the studies. Eventually, the conic solutions to all types of cubic equations were summarised by Omar Khayyam (1048 – 1131), ([6], page 80).

 

2.8 Other Mathematical Achievements

In this chapter, we will discuss some of the important, though less well-known, achievements of Archimedes.

 

2.8a Formula for the area of a triangle

Area of a triangle =, where s is the semiperimeter, a, b and c are the side lengths of the triangle.

This formula is known as the Heron’s formula. But it was in fact discovered by Archimedes ([5], page 125).

2.8b Semi-regular Polyhedron

Semi-regular polyhedron is a solid formed from faces of regular polygons with different number of sides. In addition, for identical faces, the arrangements of neighboring faces are identical. There are a total of 13 possible types of semi-regular polyhedron ([7], page 14 – 15).

Source: Reference [7], page 15

2.8c Construction of Regular Heptagon

There was a note on the construction of regular heptagon by Archimedes in some a translation work of an Arabic mathematician ([4], page 380).

2.8d The Cattle Problem

Archimedes proposed the following problem when he visited Hieron, the King of Syracuse:

Suppose W and w are the number of white bulls and cows respectively,

X and x are the number of black bulls and cows respectively,

Y and y are the number of yellow bulls and cows respectively,

Z and z are the number of dappled bulls and cows respectively.

where:

Then, what are the values of the unknowns?

Hieron spent much effort to get the answer. The smallest number got 7 digits. Just then Archimedes add two more conditions:

W + X = Perfect square
Y + Z = Triangular number

Hieron were unable to solve this. Experts on the history of mathematics believe that Archimedes himself could not solve this problem either because in 1965, with the help of computers, the answer was found to be of 206500 digits ([2], page 97 – 98).

Yet, from the solution of the first part of the problem and the difficulty of the second part, we can infer Archimedes’ ability in algebra.

2.8e “Book of Lemmas”

In the Arabic book “Book of Lemmas”, there were 15 problems on circles. These problems were believed to be reorganized by the descendents of Archimedes. The statements in “Book of Lemmas” do not seem to concur to a central theme. Neither are they done with method of exhaustion — they are just small discussions on the properties of circles. Below low are a few examples:

  1. Statement 4: If AB is the diameter of a semicircle, N an arbitrary point on AB. Then the sum of areas of the two semicircles drawn with AN and BN as diameters respectively will be equal to that of the circle with PN as radius, where PN is the perpendicular of AB from P on the semicircle.
  2. Statement 9: If AB and CD are not diameters and they intersect at right angles, then(arc AD) + (arc CB) = (arc AC) + (arc DB).

 

III. Methodologies

In his studies, Archimedes made good use of the works of predecessors, proposed innovative ideas and insisted on strict proofs ([4], page 26 – 27). As in “Elements”, his works started with a list of definitions and hypotheses. This is followed by the derivation and proofs of his statements. We will discuss the means he used in his proofs below. Besides, from “The Methods”, ([4], page 339 – 340), we know that the order of appearance of his statements does not match the chronological order of their discoveries. This means that the discoveries did not followed from the deductions in his works. We will also discuss the routes he got his ideas in this section.

 

3.1 The Use of Proportions

Archimedes used proportions in his statements and proofs. In fact, ancient Greek Mathematics mainly involved the uses of areas and proportions in solving problems. Euclid used both methods, Apollonius of Perga mainly used areas and Archimedes usually used proportions ([4], page 27). For example, Apollonius interpreted y2 as an area and pointed out the relationships of y2 = px in parabolae and in hyperbolae and ellipse, where p is the parameter and d is the diameter; however, Archimedes viewed parabolae as ([2], page 139). Hence, they came up with diverse results in their studies.

 

3.2 Method of Exhaustion and Proof by Contradiction

In the second statement of the twelfth chapter of Euclid’s “Elements”, there was a use of the method of exhaustion. Archimedes also employed this method in proving many of his statements. One example is that in 2.1.1d. Choose two arbitrary values to give a ratio R and draw regular polygons inscribing and inscribed by a circle. Archimedes proved that if the polygons have sufficient numbers of sides, the ratio between their areas must be smaller than R. Hence if R is close to 1, then the ration of the areas of the polygons is also close to 1. In this way, we can determine the area of the circle.

Circle, inscribing polygon, inscribed polygon

It was originally difficult to evaluate the area bounded by a curve. By constructing the polygons, we can make our estimate of the area arbitrarily close to the actual value. In fact, this method can be applied to the areas bounded by conics and the volumes of the solids of revolution. In “On the Sphere and Cylinder”, “On Measurement of a Circle”, “On Quadrature of the parabola”, “On Conoids and Spheroids” and “On Spirals”, Archimedes often use the method of exhaustion to solve the problems.

We need not construct both the inscribing and inscribed polygons in using the method of exhaustion: sometimes we need only one of them. One example is Statement 24 of “On Quadrature of the parabola” as mentioned before, it stated that “The area of the quadrature of any parabola is equal to times that of a triangle with the same base and height.”

The proof ([5], page 119 – 120) involves using a series of triangles to approximate the parabola. Firstly, we determine the vertex, P, of the parabola and construct DPQp. Then we find the vertices R and r between PQ and Pq, and construct DPQR and DPqr. Note that the preceding statements in the book has proved that

Area of DPQR = Area of DPqr = DPQp.

Hence, the shaded area = DPQp + DPQp.

Then he proved that if we repeat the above process in DPQR and DPqr, we can form a series of triangles such that the sume of areas differs from that of the actual segment area by an arbitrarily small value. Hence we have

Parabola PQp = DPQp + DPQp + DPQp+ DPQp+�+ DPQp

The method of exhaustion gives us the formulae for calculating areas. However, as the formulae involved infinite sequences and there was not an infinity concept, the problem was not yet completely solved. To finish the problem, we need to prove by contradiction.

In this method, we first propose a value to be the answer and hypothesized on the contrary. If we can prove that there will be a contradiction if the true value is smaller or greater than the proposed value, then we are done. Now as Archimedes proposed that the shaded area is DPQp, we may assume that

area of the parabolic segment > DPQp.

Since the difference between the area of the parabola and the sum of areas of the triangles can be made arbitrarily small, we have

area of the parabolic segment > sum of areas of the triangles > DPQp,

Yet, according to Statement 23 [Note], if we sum the series to the mth term (m > n),

sum of areas of triangles + (DPQp)= DPQp

and this would lead to a contradiction because it implies

sum of areas of triangles < DPQp.

Similarly we can prove that

area of the parabolic segment < DPQp

would also lead to a contradiction. Thus, the value of the area should be as proposed.

From the above, we can see that method of exhaustion and proof by contradiction are

 

3.3 The Use of Mechanics

Up to this point, you may become as amazed as those who studied Archimedes’ works before 1906: how could Archimedes came up with so many innovative ideas? Indeed, Archimedes’ works were so neat that one could only find his definitions, hypotheses, statements and proofs, but not his routes to gather his ideas. Now we will take a look at these ‘hidden’ routes.

In 1906, J. L. Geiberg (1854 – 1928) found in Turkey a book containing a lot of Archimedes’ works, including “The Methods”, which was written by Archimedes to describe where his beautiful ideas came from. Most of the 15 statements in the book We do not focus on the proofs here, but on something of even greater value.

Archimedes is famous for his studies in mechanics. In fact, mechanics led him to many of is discoveries in mathematics. As an example, the first statement in “The Methods” was on the areas bounded by parabolae, and the proof was discussed in the previous chapter. In the book, he wrote down how he discovered the relationship:

Figure duplicated from page 29 of Reference [2]

In the above figure, BD is the diameter of the segment, CF is the tangent at C and P is an arbitrary point on the chord of the parabola. Note that AKF and OPNM are parallel to BD, C, B, N, K and H are collinear and KH and KC have equal lengths. As B is the vertex, using the previous theorems, we have EB = BD, MN = NP, FK = KA. In Statement 5 it is also proved that

MO : OP = CA : AO = CK : KN = HK : KN

Now assume that the segments have weights proportional to the lengths. View K as the pivot of HC and put a line segment TG of equal weight as OP at H. Since the center of gravity (midpoint) of MO, N, lies on KN, the two sides of the pivot K is balanced.

Since P is arbitrary, OP can be any parallel line between the parabolic segment ABC and AC and thus MO can be any parallel line in DAFC with the aforementioned relationship with OP.

Archimedes believed that as the numbers (OP and MO) of line segments in the parabolic segment ABC and DAFC are equal, for the pivot K, the parabolic segment ABC formed from all OP will balance the DAFC formed from all MO, and the center of gravity ofDAFC is the point W on CK, where CW = 2WK.

Hence,

parabolic segment ABC: DAFC = WK : KH = 1 : 3 (∵CK = KH).

Therefore,

parabolic segment ABC = DAFC = DABC. (The previous statement proved the relationship of 4 times)

The theme of “The Methods” is to divide the unknown quantity into many small quantities and use the lever to determine the proportional relationships of these quantities with other small quantities. The sum of the latter group of quantities is smaller to calculate ( like the triangleDAFC above. This allows us to find the unknown ([6], page 74 -750). Note that this is a preliminary concept of integration, and Archimedes successfully combines the studies of mechanics and mathematics.

This method is indeed marvelous. But Archimedes pointed out that this is just a way to provide guesses, not proofs. This means that we still need other means (the method of exhaustion and prove by contradiction) to prove the results obtained. Actually, sometimes the method of using mechanics fails. Consider the figure below ([6], page 75 – 76):

DC is an altitude of the triangle. If EF // AB, EG AB and FH AB, then GE = HF. As EF goes from AB to C, at each height, we can find the corresponding EG and FH. This leads to the conclusion that the areas of DACD and DBCD are equal, which may not be true.

 

IV. Conclusion

Archimedes, Euclid and Apollonius of Perga are the three greatest mathematicians in ancient Greece. Also Archimedes is also considered one of the three greatest mathematicians ever, as with Newton and Gauss. Just like Newton, Archimedes is famous for his achievements in mechanics. Yet, his achievements in mathematics may not be neglected.

Many works were written by Archimedes, including “On the Sphere and Cylinder”, “On the Measurement of a Circle”, “On Conoids and Spheroids”, “On Spirals”, “The Sandreckoner”, “On Quadrature of the parabola”, “Book of Lemmas” and “The Methods”. These works are mainly on geometry topics: in two-dimensional plane, he worked on circles, parabolae, hyperbolae and tangents to spirals; in three-dimensional space, he discussed the areas and volumes of cylinders, cones, spheres, spheroids and conoids and the approximation of Pi. Other achievements include Heron’s formula, construction of regular heptagon, semi-regular polyhedron, and many more.

Archimedes seldom touch on the topic of arithmetic, as reflected by the fact that he had only one piece of work, “The Sandreckoner”, on the topic. He invented a new system to represent large numbers. Some of his contribution to the field arises from the need of solving geometry problems. For example, he gave the approximations of and square roots of some large numbers, insights on the solutions of cubic equations and arithmetic series.

In terms of methodologies in research, Archimedes is also outstanding. His use of proportion in his statements and mechanics in discovering relationships was pioneering. The invention of the concept of integration and the use of the method of exhaustion and proof by contradiction rendered him invaluable tools in proofs.

In conclusion, Archimedes is well deserved to be named as one of the greatest mathematicians.

 

V. References

[1] John Stillwell (1989), Mathematics and Its History, Springer.

[2] T. L. Heath (1965), A History of Greek Mathematics, Volume II, Oxford.

[3] E.T. Bell (written in 1937),Jing Zhujun (translated in 1998).Da shu xue jia (translated from Men of mathematics). Publisher: Chiu Chang Publishing Company.

[4] T. L. Heath (written in 1912), Zhu Enkuan, Li Wenming (translated in 1998). Ajimide quan ji (translated from The Works of Archimedes). Publisher: Shanxi ke xue ji shu chu ban she (Shanxi).

[5] Morris Kline (written in 1908), Lin Yanquan, Hong Wangsheng, Yang Kang Jingsong (translated in 1983). Shu xue shi : shu xue si xiang de fa zhan (translated from Mathematical thought from ancient to modern times). Publisher: Chiu Chang Publishing Company.

[6] Xie Enze, Xu Benshun (1994). Shi jie shu xue jia si xiang fang fa, p.61-99.Publisher:Shandong jiao yu chu ban she (Jinan).

[7] Cai Zhiqiang, Sun Wenxian(199?). Shu xue li ti mo xing zhi zuo : fu : duo mian ti yan jiu. Publisher: Chiu Chang Publishing Company.

[8] Hong Kong University of Science and Technology. http://www.edp.ust.hk/math/history/5/5_5/5_5_34.htm

數學資料庫 – 數學趣趣地 – 數學文章

微積分的發展

微積分是微分和積分兩門學問的統稱,研究的範疇有三,包括微分、積分,以及微分和積分兩者之間的關係。微分主要討論一個變量怎樣隨時間(或其他變量)改變,而積分則主要討論計算面積的方法。它們兩者的關係由「微積分基本定理」(或稱「牛頓 - 萊布尼茨公式」)給出:簡單來說,這條定理說明,在適當的條件下,求積分是求微分之逆,求微分也是求積分之逆。以下簡介微積分發展的歷史。

一、 萌芽時期

早在希臘時期,人類已經開始討論「無窮」、「極限」以及「無窮分割」等概念。這些都是微積分的中心思想;雖然這些討論從現代的觀點看有很多漏洞,有時現代人甚至覺得這些討論的論証和結論都很荒謬,但無可否認,這些討論是人類發展微積分的第一步。

例如公元前五世紀,希臘的德謨克利特(Democritus)提出原子論:他認為宇宙萬物是由極細的原子構成。在中國,《莊子.天下篇》中所言的「一尺之捶,日取其半,萬世不竭」,亦指零是無窮小量。這些都是最早期人類對無窮、極限等概念的原始的描述。

其他關於無窮、極限的論述,還包括芝諾(Zeno)幾個著名的悖論1:其中一個悖論說一個人永遠都追不上一隻烏龜2,因為當那人追到烏龜的出發點時,烏龜已經向前爬行了一小段路,當他再追完這一小段,烏龜又已經再向前爬行了一小段路。芝諾說這樣一追一趕的永遠重覆下去,任何人都總追不上一隻最慢的烏龜--當然,從現代的觀點看,芝諾說的實在荒謬不過;他混淆了「無限」和「無限可分」的概念。人追烏龜經過的那段路縱然無限可分,其長度卻是有限的;所以人仍然可以以有限的時間,走完這一段路。然而這些荒謬的論述,開啟了人類對無窮、極限等概念的探討,對後世發展微積分有深遠的歷史意味。

另外值得一提的是,希臘時代的阿基米德(Archimedes)已經懂得用無窮分割的方法正確地計算一些面積,這跟現代積分的觀念已經很相似。由此可見,在歷史上,積分觀念的形成比微分還要早--這跟課程上往往先討論微分再討論積分剛剛相反。

二、 十七世紀的大發展--牛頓和萊布尼茨的貢獻

中世紀時期,歐洲科學發展停滯不前,人類對無窮、極限和積分等觀念的想法都沒有甚麼突破。中世紀以後,歐洲數學和科學急速發展,微積分的觀念也於此時趨於成熟。在積分方面,一六一五年,開普勒(Kepler)把酒桶看作一個由無數圓薄片積累而成的物件,從而求出其體積。而伽利略(Galileo)的學生卡瓦列里(Cavalieri)即認為一條線由無窮多個點構成;一個面由無窮多條線構成;一個立體由無窮多個面構成。這些想法都是積分法的前驅。

在微分方面,十七世紀人類也有很大的突破。費馬(Fermat)在一封給羅貝瓦(Roberval)的信中,提及計算函數的極大值和極小值的步驟,而這實際上已相當於現代微分學中所用,設函數導數為零,然後求出函數極點的方法。另外,巴羅(Barrow)亦已經懂得透過「微分三角形」(相當於以 dxdyds 為邊的三角形)求出切線的方程,這和現今微分學中用導數求切線的方法是一樣的。由此可見,人類在十七世紀已經掌握了微分的要領。

然而,直至十七世紀中葉,人類仍然認為微分和積分是兩個獨立的觀念。就在這個時候,牛頓和萊布尼茨將微分及積分兩個貌似不相關的問題,透過「微積分基本定理」或「牛頓 - 萊布尼茨公式」連繫起來,說明求積分基本上是求微分之逆,求微分也是求積分之逆。這是微積分理論中的基石,是微積分發展一個重要的里程碑。

微積分誕生以後,逐漸發揮出它非凡的威力,過去很多初等數學束手無策的問題,至此往往迎刃而解。例如,雅各布.伯努利(Jakob Bernoulli)用微積分的技巧,發現對數螺線經過各種適當的變換之後,仍然是對數螺線3。他的弟弟約翰.伯努利(Johnann Bernoulli)在一六九六年提出一個「最速降線」問題︰「一質點受地心吸力的作用,自較高點下滑至較低點,不計摩擦,問沿著什麼曲線,時間最短?」這條問題後來促使了變分學誕生4。歐拉(Euler)的《引論》、《微分學》、《積分學》亦總結了自十七世紀微積分的全部成果。

儘管如此,微積分的理論基礎問題,仍然在當時的數學界引起很多爭論5。牛頓的「無窮小量」,有時是零,有時又不是零,他的極限理論也是十分模糊的。萊布尼茨的微積分同樣不能自圓其說。這個問題要到十九世紀才得到完滿的解答,所以微積分在當時,惹來不少反對的聲音,當中包括數學家羅爾(Rolle)。儘管如此,羅爾本身亦曾提出一條與微積分有關的定理︰他指出任意的多項式 f(x) = a + bx + cx2 + dx3 + … 的任何兩個實根之間都存在至少一個 b + 2cx + 3dx2 + … 的實根。熟悉微積分的朋友會知道,b + 2cx + 3dx2 + … 其實是 f(x) = a + bx + cx2 + dx3 + … 的導數6。後人將這條定理推廣至可微函數,發現若函數 f (x) 可微,則在 f (x) = 0 的任何兩個實根之間,方程 f’(x) = 0 至少有一個實根。這條定理被冠為「羅爾定理」,是為微分學的基本定理之一。由此可見,在挑戰微積分的理論基礎的同時,數學家已經就微積分的發展作出了很大的貢獻。

三、 十九世紀基礎的奠定

微積分的發展迅速,使人來不及檢查和鞏固微積分的理論基礎。十九世紀,許多迫切問題基本上經已解決,數學家於是轉向微積分理論的基礎重建,人類亦終於首次給出極限、微分和積分等概念的嚴格定義。

一八一六年,波爾查諾(Bolzano)在人類歷史上首次給出連續函數的近代定義。繼而在一八二一年,柯西(Cauchy)在他的《教程》中提出 e 方法,後來在一八二三年的《概要》中他改寫為 d 方法,把整個極限過程用不等式來刻畫,使無窮的運算化為一系列不等式的推算,這就是所謂極限概念的「算術化」。後來外爾斯特拉斯 (Weierstrass)將 ed 聯繫起來,完成了 ed 方法,這就是現代極限的嚴格定義。

有了極限的嚴格定義,數學家便開始嘗試嚴格定義導數和積分。在柯西之前,數學家通常以微分為微積分的基本概念,並把導數視作微分的商。然而微分的概念模糊,把導數定義作微分的商因此並不嚴謹。於是柯西《概要》中直接定義導數為差商的極限,這就是現代導數的嚴格定義,是為現代微分學的基礎。

在《概要》中,柯西還給出連續函數的積分的定義:設 f(x) 為在 [a,b] 上連續的函數,則任意用分點 a = x0 < … < xn = b,將 [a,b] 分為 n 個子區間 [xi-1,xi] (i = 1, 2, …, n),若果和式

當最大子區間的長度趨向 0 時,極限存在,則此極限稱為函數 f(x) 在 [a,b] 上的積分。這跟現代連續函數積分的定義是一致的。

後來黎曼(Riemann)推廣了柯西的定義。黎曼的定義跟柯西的定義不同的地方,在於和式 S 的定義:在黎曼的定義中,和式 S 定義為

(留意黎曼在黎曼和中用了 [xi-1,xi] 中任意一點 xi-1,而柯西在其和式 S 中則永遠選取子區間 [xi-1,xi] 的左端點 xi-1)。我們說黎曼推廣了柯西的定義,是因為對所有在 [a,b] 上連續的函數,柯西積分的值跟黎曼積分的值一樣,而且有一些在 [a,b] 上不連續的函數,當最大子區間的長度趨向 0 時 S 的極限依然存在。這就是現在所用的「黎曼積分」的定義;至此微積分理論的基礎重建已經大致完成。

柯西以後,微積分邏輯基礎發展史上的最重大事件是人類從集合理論出發,建立了實數理論--我們說實數理論的建立是微積分理論發展史上的一件大事,是因為微積分的理論用上了很多實數的性質。這實數理論的建立,主要功勞歸於戴德金(Dedekind)、康托爾(Cantor)、外爾斯特拉斯等人。一八七二年,梅雷(Méray)提出的無理數定義,和同一年康托爾提出用有理「基本序列」來定義無理數實質相同。有了實數理論,加上集合論和極限理論,微積分就自從三百年以來,首次有了鞏固的邏輯基礎,而微積分的理論亦終於趨於完備。

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每年翻開月曆時,總會發現農曆和西曆的對照年年不同。可是只要多看幾年的月曆,便會發現每隔 19 年,農曆和西曆的對照便會大致相同。

在 1984 年年尾至 1985 年年初出生的朋友,最近 19 歲生日時有沒有發現自己的農曆生日和西曆生日相差甚遠呢?我們探討這個問題的原因前,必須先知道兩種曆法的計算方式。

現時為世界普遍採用的西曆,正確名稱是「儒略曆」(Gregorian calendar)。這套曆法的雛形由古羅馬凱撒大帝 (Julius Caesar) 在公元前 46 年修訂當時的曆法制定,後來再在 1576 年由羅馬教皇再修訂成現代的曆法(因為 1576 年的修訂原因和本題目無關,這裏略過不談)。因為這套曆法的根據是地球環繞太陽公轉的周期(365.2422天),所以它可被歸類為「陽曆」(solar calendar)。曆法內容如下:

1. 閏年

當年份數字是 4 的倍數時,該年則為「閏年」,否則為「平年」;但若年份為 100 的倍數而不是 400 的倍數,該年則改回「平年」。

2. 每月的日數

平年每年共分為 365 天,閏年則為 366 天,這三百幾天約略平均分成 12 個月,其中 1、3、5、7、8、10 和 12 月有 31 天,4、6、9 和 11 月有 30 天,平年的 2 月有 28 天,閏年的 2 月則有 29 天。(這種分配方法有其歷史原因,但這裏不作深究。)

簡單來說,每 400 年共有 97 個閏年,即 365 ´ 400 + 97 = 146097 天,這和 400 年共有 365.2422 ´ 400 = 146096.88 天 只差 0.12 天,誤差很小。因此,它可以準確地量度地球公轉的活動。

我們指的農曆是中國農曆,應始用於中國夏朝。商朝時農曆已有相當準確的曆法計算規則,現代的農曆大部分規則亦建基於此。農曆的計算依據比儒略曆複雜。平日我們說它是陰曆 (lunar calendar),其實是一種誤解。它既考慮地球的公轉周期,亦考慮月亮環繞地球公轉的周期,因此它被稱為「陰陽曆」(lunisolar calendar)。實際上真正的陰曆不多,其中回教曆 (Islamic calendar) 即為陰曆。[1, 2]

1. 初一的位置

計算農曆時,我們首先計算每月的初一是在哪一天。眾所周知,初一時月球亮度最小,是新月開始的日子。因此,我們會將新月開始的一天定為初一(月份序號尚未確定)1。因為月亮的盈虧周期是 29.5306 日,所以農曆每個月都是 29 或 30 日。

2. 十一月的位置

決定月份的序號前,我們應先知道節氣和「歲」的概念。

我們將地球公轉軌道分為 24 份,每份為 15 度,代表一個節氣。當地球到達進入某份的軌道時,當天便屬於該份的節氣。二十四個節氣順序為

1. 立春 2. 雨水 3. 驚蟄 4. 春分 5. 清明 6. 穀雨
7. 立夏 8. 小滿 9. 芒種 10. 夏至 11. 小暑 12. 大暑
13. 立秋 14. 處暑 15. 白露 16. 秋分 17. 寒露 18. 霜降
19. 立冬 20. 小雪 21. 大雪 22. 冬至 23. 小寒 24. 大寒

表 1:二十四節氣表

其中第偶數個的節氣(如雨水、春分、……、冬至和大寒)都稱為「中氣」。

每個「歲」是由十一月開始,翌年十一月前的一個月結束。根據規定,冬至一定要在每個「歲」的首月。因此,我們會將包含冬至的那個月定為十一月。

為甚麼人們以十一月而不是正月為計算基礎,又以冬至為算法的準則呢?首先,對位於北半球的中國來說,冬至是全年影子最長的一天2。因此,古人只需利用垂直的竹竿測影便可得知哪天是冬至。至於以十一月為計算基礎則為古代的傳統。這種曆法安排始於夏朝,幾經改動後在西漢再度採用,並沿用至今3。另外,這亦和中國農民生活習慣有關。冬至是農民開始休息的日子。從冬至至立春,人們一般都會與家人準備團年,不會耕作,讓農地休養生息。因此,冬至對農民來說是重要的時間標記。這就是計算曆法的人以冬至作為算法準則的原因。「歲」的重要性亦可從中國古代計算年齡的方法得知。中國古代的人數算年份時,往往會以歲4為數算的基礎。例如,人們計算自己的出生年份時,他們會考慮自己是在哪一「歲」出生,而不是哪一「年」出生。

3. 節氣和閏月

其他的月份序號怎樣決定呢?在此之前,我們會先決定閏月。閏月的規則如下:

1. 如果某個歲有 13 個月,則該歲為閏歲。

2. 如果某歲不是閏歲,月份則按著十一月、十二月、一月、二月、……、十月順序編號。

3. 如果某歲是閏歲,我們便要將其中一個月定為閏月。決定閏月的規則和節氣有關:一年有 12 個中氣。因此,13 個月裏必定有最少一個月沒有中氣5。閏歲中第一個沒有中氣的月份是該歲的閏月。除了閏月的月份序號和前一個月一樣外,其他月份都按著十一月、十二月、一月、二月、……、十月順序編號。

由於 19 年共有 365.2422 ´ 19 = 6936.6018 日,而 19 ´ 12 + 7 = 235 個農曆月則有 235 ´ 29.5306 = 6936.6910 日,兩者僅相差約 0.08 天。故若 19 年置 7 閏,農曆便可以頗準確地描述地球公轉的情況。這便是人們常說「十九年七閏」的原因。這亦解釋了為何農曆和儒略曆的日期對照會每 19 年便「循環」一次6[1,4]。

按著上述的周期推斷,1984/1985 年的農曆對照應該和 2003/2004 年的差不多吧。可是它們的閏月位置卻不相同。問題在哪裏呢?原來問題是在 1984/85 年,不在 2003/04 年。1984 年(甲子)有閏十月,1985 年不閏;可是 2003 年卻沒有閏年,2004 年則有閏二月。為甚麼呢?這正和那些年份是否閏歲有關。我們先看看 1984、1985、2003 和 2004 部分節氣和新月的資料[3, 5]。

1984/85 年 2003/04 年
節令 日期及時間 備註 節令 日期及時間 備註
小雪 22/11/1984 (11:11) 十月 小雪 23/11/2003 (01:31) 十月
新月 23/11/1984 (06:58) 此月沒有中氣,故這是閏十月 新月 24/11/2003 (07:00) 此月有冬至,故這是十一月
大雪 07/12/1984 (06:24) 大雪 07/12/2003 (20:55)
新月 22/12/1984 (19:47) 此月有冬至,故這是十一月 冬至 22/12/2003 (14:44)
冬至 22/12/1984 (23:06) 新月 23/12/2003 (17:44) 十二月
小寒 05/01/1985 (17:33) 小寒 06/01/2004 (08:04)
大寒 20/01/1985 (10:44) 大寒 21/01/2004 (01:22)
新月 21/01/1985 (10:29) 十二月 新月 22/01/2004 (05:06) 一月
立春 04/02/1985 (05:15) 立春 04/02/2004 (19:46)
雨水 19/02/1985 (00:14) 雨水 19/02/2004 (15:39)
新月 20/02/1985 (02:43) 正月 新月 20/02/2004 (17:18) 二月
驚蟄 05/03/1985 (23:25) 驚蟄 05/03/2004 (13:56)
新月 21/03/1985 (19:59) 二月 春分 20/03/2004 (14:52)
春分 21/03/1985 (00:21) 新月 21/03/2004 (06:42) 此月沒有中氣,故這是閏二月
清明 05/04/1985 (04:28) 清明 04/04/2004 (18:44)
新月 20/04/1985 (13:22) 三月 新月 19/04/2004 (21:22) 三月
穀雨 20/04/1985 (11:34) 穀雨 20/04/2004 (02:11)

表 2:1984/85 及 2003/04 年的部分節氣及新月資料(綠色的是中氣,藍色的不是中氣)

從表 2 得知,1984 年的冬至當天剛好是新月開始,所以 1984 年 12 月 22 日開始的那個月是十一月。但 2003 年的冬至卻在新月開始前一天。因此,2003 年 12 月 23 日開始的那個月是十二月。正因這個小差別,1984 年的歲有 13 個月,是閏歲;但 2003 年的歲卻只有 12 個月,不是閏歲。因此,1984 年便有閏十月,2003 年沒有閏月。相似地,1985 年的歲有 12 個月但 2004 年的歲卻有 13 個,所以 1985 年沒有閏月,2004 年有閏二月。

這個異常的閏月情況令 1984 年 11 月 23 日至 1985 年 4 月 19 日之間出生的人的 19 歲農曆生日和儒略曆生日相差甚遠,而且這兩個生日在未來一千年都不可能重疊。以三個在 1984 和 1985 年出生的人為例子,他們幾次生日的資料如下:

儒略曆生日時的農曆對照 1984 年 10 月 14 日
(農曆為九月二十日)
1985 年 3 月 14 日
(農曆為正月二十三日)
1985 年 9 月 10 日
(農曆為七月二十六日)
19 歲生日 九月十九日 二月二十四日 七月二十六日
38 歲生日 九月十九日 二月二十三日 七月二十六日
57 歲生日 九月二十日 二月二十三日 七月二十六日
76 歲生日 九月二十一日 二月二十三日 七月二十七日
95 歲生日 九月二十日 二月二十三日 七月二十七日

表 3:三人的儒略曆生日的農曆對照

從表 3 可見,不在上述時段出生的人在 19、38、57、76 和 95 歲的生日中總有幾次是儒略曆生日及農曆生日重疊。可是 1985 年 3 月 14 日出生的人卻一次也沒有,這就是閏月不每 19 年出現的異常情況所致。事實上,這種異常情況並不常見。上一次發生這種情況是 1889/90 年,而下一次則在 2185/86 年(1890 年和 2186 年都有閏二月)。[1]

[1] Helmer Aslaksen, The Mathematics of the Chinese Calendar, 2000

[2] Helmer Aslaksen, An Introduction to the Chinese Calendar, 2000

[3] 鄧匡哲、渡森、觀龍、黎逢華合編,標準中西對照萬年曆,玄學出版社

[4] KUAN Shau Hong and TENG Keat Huat, The Chinese Calendar of The Later Han Period,1999

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Logarithm and Logarithm Table

Introduction

Logarithm tables are commonly listed in the appendices of Mathematics textbooks to ease readers’ calculations. However, do you know it was before the invention of the concept of logarithm that the first logarithm table was published?

Background

In the fifteenth century, the Renaissance in Europe sped up the development of distant transport. Yet, at that time, it was a tedious work to perform multiplications and divisions on the large numbers involved in distant transport. And there arouse a question: Does there exist an accurate and efficient way to perform arithmetic operations on large numbers?

Replacing Multiplication and Division with Addition and Subtraction

In order to simplify the calculations, a natural thought was to replace multiplication and division with addition and subtraction.

This idea was illustrated by two transformation formulae emerging in the sixteenth century:

However, both formulae were not the perfect answers to the question: the former required manipulations of sine and cosine ratios in trigonometry, while the latter needed evaluations of squares of the numbers, and this would involve some tedious multiplications.

A Discovery

In 1484, Nicolas Chuquet, a bachelor of medicine in Paris, discovered an interesting mathematical phenomenon:

2 + 4 = 6

The first row is an arithmetic sequence with a common difference of 1, and the second row is a geometric sequence with a common ratio of 2.

He discovered that the product of any two terms in the geometric sequence lies in the sequence, and the sum of the two corresponding terms in the arithmetic sequence corresponds to this product too.

For example, consider the terms 4 and 16 in the geometric sequence. Their product, 64, is a term in the same sequence. Besides, as the number above 4 ´ 16 = 64 is 6, this product corresponds to the number 6, which is the sum of the two terms (2 and 4) corresponding to 4 and 16 in the sequence.

Chuquet’s discovery was inspiring. The implication was that one can convert multiplication to addition by a correspondence between an arithmetic sequence and a geometric sequence.

The Embryo of the Logarithm Table

Half a century later, a German Mathematician named Michael Stifel introduced the use of negative numbers and rational numbers in geometric and arithmetic sequences. He went further to point out that multiplication and division in geometric sequences can be converted to addition and subtraction in arithmetic sequences respectively.

For example, to evaluate , one may notice that 5 is the number corresponding to 32 in the arithmetic sequence and -1 is that corresponding to . As 5 – (-1) = 6 and 6 corresponds to 64, the answer is 64.

5 – (-1) = 6

The Invention of the Logarithm Table

The first person to give a definition to logarithm and propose the logarithm table is John Napier, a Scottish mathematician. He defined logarithm with the help of physical motions.

He assumed that there are two particles P and Q traveling along the rays AZ and A’Z’ respectively. P maintains a constant velocity, while the velocity of Q at any point is proportional to the remaining distance (this means that the velocity of Q is decreasing).

Suppose that when P gets to B’, Q is at B, and when P gets to C’, D’, E’, …, Q is at C, D, E, ….Then, the distance A’B’ is equal to the logarithm of the distance BZ. Similarly, distances A’C’, A’D’ and A’E’ are equal to the logarithms of the distances CZ, DZ and EZ respectively.

Q:
P:

In this way, Napier defined the logarithmic values of the numbers in a geometric sequence to be the corresponding numbers in the corresponding arithmetic sequence. Hence, the logarithmic values defined by Napier with physical motions or geometry are continuous, while those defined with sequences are discrete. Of course, even Napier could not list out the logarithm of every number in the sequence when he constructed his logarithm table.

Napier spent nearly 20 years to construct his logarithm table to a precision of 7 significant figures.

A More Precise Logarithm Table

The publication of Napier’s logarithm table was fully supported by the British Mathematician Henry Briggs, who then suggested that Napier should use base 10 in his logarithm table for the sake of convenience in the decimal system.

While Napier agreed to Briggs’s advice, he was too tired to finish off the work. To accomplish the goal, Briggs and a young Dutch mathematician Adriaan Vlacq constructed the common logarithm table, to the precision of 14 significant figures, for the numbers between 1 and 100000.

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52. 6th Pui Ching Invitational Mathematics Competition (Final, Answers) 20 / 03 / 2007 Chi / Eng PCMS/PCEC
53. 7th Pui Ching Invitational Mathematics Competition (Heat, Secondary 1) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
54. 7th Pui Ching Invitational Mathematics Competition (Heat, Secondary 2) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
55. 7th Pui Ching Invitational Mathematics Competition (Heat, Secondary 3) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
56. 7th Pui Ching Invitational Mathematics Competition (Heat, Secondary 4) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
57. 7th Pui Ching Invitational Mathematics Competition (Heat, Senior) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
58. 7th Pui Ching Invitational Mathematics Competition (Heat, Answers) 14 / 02 / 2008 Chi / Eng PCMS/PCEC
59. 7th Pui Ching Invitational Mathematics Competition (Final, Secondary 1) 23 / 09 / 2008 Chi / Eng PCMS/PCEC
60. 7th Pui Ching Invitational Mathematics Competition (Final, Secondary 2) 23 / 09 / 2008 Chi / Eng PCMS/PCEC
61. 7th Pui Ching Invitational Mathematics Competition (Final, Secondary 3) 23 / 09 / 2008 Chi / Eng PCMS/PCEC
62. 7th Pui Ching Invitational Mathematics Competition (Final, Secondary 4) 23 / 09 / 2008 Chi / Eng PCMS/PCEC 63. 7th Pui Ching Invitational Mathematics Competition (Final, Senior) 23 / 09 / 2008 Chi / Eng PCMS/PCEC 64. 7th Pui Ching Invitational Mathematics Competition (Final, Answers) 23 / 09 / 2008 Chi / Eng PCMS/PCEC
65. 8th Pui Ching Invitational Mathematics Competition (Heat, Secondary 1) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
66. 8th Pui Ching Invitational Mathematics Competition (Heat, Secondary 2) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
67. 8th Pui Ching Invitational Mathematics Competition (Heat, Secondary 3) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
68. 8th Pui Ching Invitational Mathematics Competition (Heat, Secondary 4) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
69. 8th Pui Ching Invitational Mathematics Competition (Heat, Senior) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
70. 8th Pui Ching Invitational Mathematics Competition (Heat, Answers) 25 / 01 / 2009 Chi / Eng PCMS/PCEC
71. 8th Pui Ching Invitational Mathematics Competition (Final, Secondary 1) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
72. 8th Pui Ching Invitational Mathematics Competition (Final, Secondary 2) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
73. 8th Pui Ching Invitational Mathematics Competition (Final, Secondary 3) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
74. 8th Pui Ching Invitational Mathematics Competition (Final, Secondary 4) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
75. 8th Pui Ching Invitational Mathematics Competition (Final, Secondary 5) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
76. 8th Pui Ching Invitational Mathematics Competition (Final, Answers) 15 / 03 / 2009 Chi / Eng PCMS/PCEC
77. 9th Pui Ching Invitational Mathematics Competition (Heat, Secondary 1) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
78. 9th Pui Ching Invitational Mathematics Competition (Heat, Secondary 2) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
79. 9th Pui Ching Invitational Mathematics Competition (Heat, Secondary 3) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
80. 9th Pui Ching Invitational Mathematics Competition (Heat, Secondary 4) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
81. 9th Pui Ching Invitational Mathematics Competition (Heat, Senior) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
82. 9th Pui Ching Invitational Mathematics Competition (Heat, Answers) 25 / 02 / 2010 Chi / Eng PCMS/PCEC
83. 9th Pui Ching Invitational Mathematics Competition (Final, Secondary 1) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
84. 9th Pui Ching Invitational Mathematics Competition (Final, Secondary 2) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
85. 9th Pui Ching Invitational Mathematics Competition (Final, Secondary 3) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
86. 9th Pui Ching Invitational Mathematics Competition (Final, Secondary 4) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
87. 9th Pui Ching Invitational Mathematics Competition (Final, Secondary 5) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
88. 9th Pui Ching Invitational Mathematics Competition (Final, Answers) 20 / 03 / 2010 Chi / Eng PCMS/PCEC
89. 10th Pui Ching Invitational Mathematics Competition (Heat, Secondary 1) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
90. 10th Pui Ching Invitational Mathematics Competition (Heat, Secondary 2) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
91. 10th Pui Ching Invitational Mathematics Competition (Heat, Secondary 3) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
92. 10th Pui Ching Invitational Mathematics Competition (Heat, Secondary 4) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
93. 10th Pui Ching Invitational Mathematics Competition (Heat, Senior) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
94. 10th Pui Ching Invitational Mathematics Competition (Heat, Answers) 08 / 02 / 2011 Chi / Eng PCMS/PCEC
95. 10th Pui Ching Invitational Mathematics Competition (Final, Secondary 1) 08 / 07 / 2011 Chi / Eng PCMS/PCEC
96. 10th Pui Ching Invitational Mathematics Competition (Final, Secondary 2) 08 / 07 / 2011 Chi / Eng PCMS/PCEC
97. 10th Pui Ching Invitational Mathematics Competition (Final, Secondary 3) 08 / 07 / 2011 Chi / Eng PCMS/PCEC
98. 10th Pui Ching Invitational Mathematics Competition (Final, Secondary 4) 08 / 07 / 2011 Chi / Eng PCMS/PCEC
99. 10th Pui Ching Invitational Mathematics Competition (Final, Secondary 5) 08 / 07 / 2011 Chi / Eng PCMS/PCEC
100. 10th Pui Ching Invitational Mathematics Competition (Final, Answers) 08 / 07 / 2011 Chi / Eng PCMS/PCEC

Mathematical Database – Teaching Module – Iterations, Fractals and Chaos

Teaching Module: Iterations, Fractals and Chaos

Target

Time spent

Brief Description

Content

Teachers’ Guide

 

Target

Form 4-7 students who are interested in mathematics.

Time spent

1.25 hours to 1.5 hours for each lesson.

Brief Description

In this module we aim to guide students to explore, under a unified framework, some phenomena concerning iterations, fractals and chaos. We emphasize concrete examples, and we believe that it is through working out concrete computations that students can have a better feeling of the mathematics behind these deep subjects. Therefore we have designed a series of student-oriented mathematical experiments throughout this module.

The module consists of a total of 7 lessons. In lessons 1 and 2 we introduce various examples of fractals including the Sierpinski Triangle, the Cantor Middle Thirds and the Snowflake. It will be seen how iterations of simple functions lead to complicated geometric objects like fractals. In lesson 3 the logistic map f (x) = ax(1-x) is introduced as a mathematical model of population growth. After that students are led to explore the iterations of this logistic map, and they will be asked to give an interpretation of the observed phenomenon in terms of population growth. Chaos will be observed in lesson 4 via the exploration of the logistic map, and some common misconceptions about chaos will be clarified. Lesson 5 presents a more theoretical explanation about some phenomena observed when we study the iterations of the logistic map in lesson 3. In particular the concept of stability will be emphasized. Towards the end of lesson 5 we guide students to study the iterates of the real quadratic map f (x) = x2 + c, which is a close relative of the logistic map. Then lesson 6 studies the iterates of the complex quadratic map, where students will meet the famous fractals: the Mandelbrot Set and Julia Sets. Finally we conclude the module with lesson 7, where students are led to explore the Newton’s method from an iteration point of view. Fractals again naturally arise when we solve equations iteratively using the complex Newton’s method. We hope finally students will appreciate how fractals, iterations and chaos come together naturally as a coherent whole.

Content

Teachers’ Guide

The teachers’ guide of this teaching module is avaliable here.

Mathematical Database – Teaching Module – Iterations, Fractals and Chaos – Lesson 6

Teaching Module (Iterations, Fractals and Chaos)

Lesson 6

Title: Fractals again — the Mandelbrot Set and the Julia Sets

Objective: To investigate the famous fractals, the Mandelbrot Set and the Julia Sets, via iterations of the complex quadratic map.

Before the lesson,

  • Prepare one copy of worksheet for each student.
Worksheet Download PDF View all GIF View page by page

Note: it is highly recommended to use PDF to print the worksheet.